Arithmetic, number systems, percentages, ratios, profit and loss, time and work, algebra, geometry, mensuration, and trigonometry.
What should come in place of the question mark '?' in the following number series?
10, 18, 34, ?, 130, 258
32
60
66
68
None of these
Correct answer
66
Why this answer is correct
Calculation:
The series follows following pattern:
⇒ 10 + (1 × 8) = 18
⇒ 18 + (2 × 8) = 34
⇒ 34 + (4 × 8) = 66
⇒ 66 + (8 × 8) = 130
∴ The value of ? is 66
What should come in place of the question mark ‘?’ in the following number series?
144, 164, 194, 234, 284, ?
400
344
564
356
369
Correct answer
344
Why this answer is correct
Calculation:
The series follows following pattern:
⇒ 144 + 20 = 164
⇒ 164 + 30 = 194
⇒ 194 + 40 = 234
⇒ 234 + 50 = 284
By observing the pattern, we can write
284 + 60 = 344
⇒ ? = 344
∴ The value of ? is 344.
What should come in place of question mark ‘?’ in the following number series?
500, 475, 426, 345, 224, ?
51
52
53
54
55
Correct answer
55
Why this answer is correct
Calculation:
Considering the above series,
500, 475, 426, 345, 224, ?
The logic of the series can be explained as,
500 – (5)2 = 475
475 – (7)2 = 426
426 – (9)2 = 345
345 – (11)2 = 224
By observing the pattern, we can write
224 – (13)2 = 55
⇒ ? = 55
∴ The value of ? is 55
What should come in place of the question mark (?) in the following number series?
33, 321, 465, 537, 573, ?
600
591
585
498
None of these
Correct answer
591
Why this answer is correct
Calculation∶
The series follows the following pattern:
⇒ 33 + 288 = 321
⇒ 321 + 144 = 465 , [ 288/2=144]
⇒ 465 + 72 = 537 ,[144/2=72]
⇒ 537 + 36 = 573 ,[72/2=36]
⇒ 573 + 18 = 591 ,[36/2=18]
∴ The value of ? is 591
Note:
Best way to approach this type of question is to find the difference from the end where gap between the number is less. here it can be 573 and 537.
A man borrowed Rs.X at a simple interest of 35% per annum for 6 years and Rs.Y at a simple interest of 20% per annum for 6 years. If he had borrowed Rs. (X + Y) at a simple interest of 25% for 6 years, he would have paid the same interest. Find the ratio of X and Y.
2 : 3
1 : 2
3 : 4
3 : 5
2 : 5
Correct answer
1 : 2
Why this answer is correct
Shortcut Trick
Apply Alligation Method on the interest rates since the time period (6 years) is constant for all parts.
Rate 1 (X) = 35%, Rate 2 (Y) = 20%, Combined Rate (X+Y) = 25%.
Difference 1 (X - Combined) = 35 − 25 = 10.
Difference 2 (Combined - Y) = 25 − 20 = 5.
Ratio X : Y = 5 : 10 = 1 : 2.
∴ The correct answer is 1 : 2.
Alternate Method
Given: Principal X at 35%, Principal Y at 20%, Combined Principal (X+Y) at 25%, Time = 6 years.
Formula Used: SI = (P × R × T) ÷ 100
⇒ Total SI = SIX + SIY
⇒ [(X × 35 × 6) ÷ 100] + [(Y × 20 × 6) ÷ 100] = [(X + Y) × 25 × 6] ÷ 100
⇒ 35X + 20Y = 25(X + Y)
⇒ 35X + 20Y = 25X + 25Y
⇒ 35X − 25X = 25Y − 20Y
⇒ 10X = 5Y
⇒ X ÷ Y = 5 ÷ 10
⇒ X : Y = 1 : 2
∴ The correct answer is 1 : 2.
Additional Information
Simple Interest Property
When the time period is constant, the total interest earned is the sum of interest from individual parts: SItotal = SI1 + SI2.
Alligation in SI
Alligation can be used to find the ratio of investments (X and Y) when the overall effective rate of interest is known for a common time period.
Relationship of R and P
For a fixed SI and Time, the Principal and Rate are inversely proportional: P1 × R1 = P2 × R2.
Two people A and B got different marks in 3 different subjects’ viz. Mathematics, English, and Science. If the total marks obtained by A is 20% more than the total marks of B and average marks obtained by B in all three subjects is 70 out of 100 and If A got 80 marks in science and 95 marks in Mathematics, then find the marks of A in English.
75
77
80
67
87
Correct answer
77
Why this answer is correct
Shortcut Trick
Total marks of B = 3 × 70 = 210.
Total marks of A = 1.20 × 210 = 252.
Marks of A in English = 252 − (80 + 95) = 252 − 175 = 77.
∴ The correct answer is 77 marks.
Alternate Method
Given:
Average marks of B in 3 subjects = 70
Total marks of A = 120% of B's total marks
Marks of A in Science = 80, Mathematics = 95
Calculations:
⇒ Total marks of B = 70 × 3 = 210
⇒ Total marks of A = 210 × (120/100) = 210 × 1.2 = 252
⇒ Total marks of A = Marks in (Maths + Science + English)
⇒ 252 = 95 + 80 + Marks in English
⇒ 252 = 175 + Marks in English
⇒ Marks in English = 252 − 175 = 77
∴ The correct answer is 77 marks.
Additional Information
Averages
Average = (Sum of all observations) ÷ (Total number of observations). Sum = Average × Count.
Percentage Increase
To increase a value by x%, multiply the original value by (100 + x)/100.
A, B, and C are partners. A invested Rs. 20,000 in business for 't' months. B invested Rs. 30,000 for 4 months less than the period invested by A and C invested Rs. 60,000 for 2 months less than the period invested by B. Their profit sharing ratio is 2 : 2 : 3. Find the number of months for which each partner has invested in the business respectively.
12, 8 and 6 months
12, 6 and 6 months
12, 7 and 5 months
12, 8 and 3 months
12, 9 and 7 months
Correct answer
12, 8 and 6 months
Why this answer is correct
Shortcut Trick
Ratio of Investment × Time equals the Profit Ratio.
For A and B: (20,000 × t) ÷ (30,000 × (t − 4)) = 2 ÷ 2.
Simplified: 2t ÷ 3(t − 4) = 1 ⇒ 2t = 3t − 12 ⇒ t = 12 months.
Time for B = t − 4 = 12 − 4 = 8 months.
Time for C = (t − 4) − 2 = 8 − 2 = 6 months.
∴ The correct answer is 12, 8 and 6 months.
Alternate Method
Given: Investments of A, B, and C are Rs. 20,000, Rs. 30,000, and Rs. 60,000. Profit ratio is 2 : 2 : 3.
Formula: Profit Ratio = (I1 × T1) : (I2 × T2) : (I3 × T3)
⇒ Let time for A be t. Then B is t − 4 and C is (t − 4) − 2 = t − 6.
⇒ Ratio of Profit (A:B) = (20,000 × t) : (30,000 × (t − 4)) = 2 : 2.
⇒ 2t ÷ (3t − 12) = 1 ⇒ 2t = 3t − 12 ⇒ t = 12.
⇒ Time for A = 12 months, B = 8 months, C = 6 months.
⇒ Check with C: (30,000 × 8) : (60,000 × 6) = 240,000 : 360,000 = 2 : 3 (Matches).
∴ The correct answer is 12, 8 and 6 months.
Additional Information
Partnership Ratios
In a partnership, profit is distributed based on the product of Capital and Time. If capitals are C1, C2 and times are T1, T2, then Profit Ratio P1:P2 = C1T1 : C2T2.
Time Calculation
If profit ratios and capitals are known, Time ratio can be found using T = P ÷ C.
Directions: The following table shows the average number of students (Boys + Girls) got graduation and fractional value of the number of girls who got graduation in four different colleges P, Q, R and S.
| Departments | Average number of (Boys + Girls) got graduation | Fractional value of number of girls got graduation |
|---|---|---|
| P | 1.5x | 2/5 |
| Q | 250% of x | 2/5 |
| R | 400 | 3/8 |
| S | 2x | 1/2 |
Note:
Average number of girls got graduation in all the given colleges = 335
If 3/5 of the boys graduated from department Q graduated with distinction and the remaining are first class and 5/8 of the girls from department Q got graduated with first class. Find the total number of students who graduated with the first class in department Q.
430
530
490
320
410
Correct answer
490
Why this answer is correct
General solution:
[2 × 1.5x × (2/5) +2 × 2.5x × (2/5) + 2 × 400 × (3/8) + 2 × 2x × (1/2)] = 1340
[1.2x + 2x + 300 + 2x] = 1340
5.2x = 1040
x = 200
Department P:
Total students (Boys + Girls) got graduated = 1.5 × 2 × 200 = 600
Number of girls got graduated = 600 × (2/5)= 240
Number of boys got graduated = (600 - 240) = 360
Department Q:
Total students (Boys + Girls) got graduated = (250 × x/100) = 2.5 × 2 × 200 = 1000
Number of girls got graduated = 1000 × (2/5)= 400
Number of boys got graduated = (1000 - 400) = 600
Department R:
Total students (Boys + Girls) got graduated = 400 × 2 = 800
Number of girls got graduated = 800 × (3/8)= 300
Number of boys got graduated = (800 - 300) = 500
Department S:
Total students (Boys + Girls) got graduated = 2 × 2 × 200 = 800
Number of girls got graduated = 800 × (1/2) = 400
Number of boys got graduated = (800 - 400) = 400
| Departments | Number of boys graduated | Number of girls graduated |
|---|---|---|
| P | 360 | 240 |
| Q | 600 | 400 |
| R | 500 | 300 |
| S | 400 | 400 |
Solution:
Number of boys graduated form department Q = 600
Number of boys graduated from department Q with distinction = 600 × 3/5 = 360
Number of boys graduated from department Q with first class = (600 - 360) = 240
Number of girls graduated from department Q with first class = (400 × 5/8) = 250
Required total = (240 + 250) = 490
Directions: The following table shows the average number of students (Boys + Girls) got graduation and fractional value of the number of girls who got graduation in four different colleges P, Q, R and S.
| Departments | Average number of (Boys + Girls) got graduation | Fractional value of number of girls got graduation |
|---|---|---|
| P | 1.5x | 2/5 |
| Q | 250% of x | 2/5 |
| R | 400 | 3/8 |
| S | 2x | 1/2 |
Note:
Average number of girls got graduation in all the given colleges = 335
If 30% and 40% of the boys graduated in department Q and R respectively are hostelers and remaining are day scholars. Total hostelers both boys and girls, graduated in department Q and R is 600. Find the number of girls graduated in department Q and R, from the hostel.
280
260
230
220
240
Correct answer
220
Why this answer is correct
General solution:
[2 × 1.5x × (2/5) +2 × 2.5x × (2/5) + 2 × 400 × (3/8) + 2 × 2x × (1/2)] = 1340
[1.2x + 2x + 300 + 2x] = 1340
5.2x = 1040
x = 200
Department P:
Total students (Boys + Girls) got graduated = 1.5 × 2 × 200 = 600
Number of girls got graduated = 600 × (2/5)= 240
Number of boys got graduated = (600 - 240) = 360
Department Q:
Total students (Boys + Girls) got graduated = (250 × x/100) = 2.5 × 2 × 200 = 1000
Number of girls got graduated = 1000 × (2/5)= 400
Number of boys got graduated = (1000 - 400) = 600
Department R:
Total students (Boys + Girls) got graduated = 400 × 2 = 800
Number of girls got graduated = 800 × (3/8)= 300
Number of boys got graduated = (800 - 300) = 500
Department S:
Total students (Boys + Girls) got graduated = 2 × 2 × 200 = 800
Number of girls got graduated = 800 × (1/2) = 400
Number of boys got graduated = (800 - 400) = 400
| Departments | Number of boys graduated | Number of girls graduated |
|---|---|---|
| P | 360 | 240 |
| Q | 600 | 400 |
| R | 500 | 300 |
| S | 400 | 400 |
Solution:
Number of boys graduated in department Q from hostel = 600 × (30/100) = 180
Number of girls graduated in department R from hostel = 500 × (40/100) = 200
Total hostelers graduated from department Q and R = 600
Number of girls graduated in department Q and R, from the hostel = 600 - (200 + 180) = 220
Directions: The following table shows the average number of students (Boys + Girls) got graduation and fractional value of the number of girls who got graduation in four different colleges P, Q, R and S.
| Departments | Average number of (Boys + Girls) got graduation | Fractional value of number of girls got graduation |
|---|---|---|
| P | 1.5x | 2/5 |
| Q | 250% of x | 2/5 |
| R | 400 | 3/8 |
| S | 2x | 1/2 |
Note:
Average number of girls got graduation in all the given colleges = 335
Number of boys who graduated from department Q has increased by 20%. The number of boys who graduated from department Q is what percentage of the number of girls who graduated from all the four departments.
48.89%
41.90%
53.73%
43.78%
55.78%
Correct answer
53.73%
Why this answer is correct
General solution:
[2 × 1.5x × (2/5) +2 × 2.5x × (2/5) + 2 × 400 × (3/8) + 2 × 2x × (1/2)] = 1340
[1.2x + 2x + 300 + 2x] = 1340
5.2x = 1040
x = 200
Department P:
Total students (Boys + Girls) got graduated = 1.5 × 2 × 200 = 600
Number of girls got graduated = 600 × (2/5)= 240
Number of boys got graduated = (600 - 240) = 360
Department Q:
Total students (Boys + Girls) got graduated = (250 × x/100) = 2.5 × 2 × 200 = 1000
Number of girls got graduated = 1000 × (2/5)= 400
Number of boys got graduated = (1000 - 400) = 600
Department R:
Total students (Boys + Girls) got graduated = 400 × 2 = 800
Number of girls got graduated = 800 × (3/8)= 300
Number of boys got graduated = (800 - 300) = 500
Department S:
Total students (Boys + Girls) got graduated = 2 × 2 × 200 = 800
Number of girls got graduated = 800 × (1/2) = 400
Number of boys got graduated = (800 - 400) = 400
| Departments | Number of boys graduated | Number of girls graduated |
|---|---|---|
| P | 360 | 240 |
| Q | 600 | 400 |
| R | 500 | 300 |
| S | 400 | 400 |
Solution:
Number of boys graduated from department Q = 600 × 120/100 = 720
Total number of girls graduated from all the department = (240 + 300 + 400 + 400) = 1340
Required percentage = (720/1340) × 100 = 53.73%
Directions: The following table shows the average number of students (Boys + Girls) got graduation and fractional value of the number of girls who got graduation in four different colleges P, Q, R and S.
| Departments | Average number of (Boys + Girls) got graduation | Fractional value of number of girls got graduation |
|---|---|---|
| P | 1.5x | 2/5 |
| Q | 250% of x | 2/5 |
| R | 400 | 3/8 |
| S | 2x | 1/2 |
Note:
Average number of girls got graduation in all the given colleges = 335
Find the difference between the average number of boys and girls graduated from all the given departments.
150
140
130
120
110
Correct answer
130
Why this answer is correct
General solution:
[2 × 1.5x × (2/5) +2 × 2.5x × (2/5) + 2 × 400 × (3/8) + 2 × 2x × (1/2)] = 1340
[1.2x + 2x + 300 + 2x] = 1340
5.2x = 1040
x = 200
Department P:
Total students (Boys + Girls) got graduated = 1.5 × 2 × 200 = 600
Number of girls got graduated = 600 × (2/5)= 240
Number of boys got graduated = (600 - 240) = 360
Department Q:
Total students (Boys + Girls) got graduated = (250 × x/100) = 2.5 × 2 × 200 = 1000
Number of girls got graduated = 1000 × (2/5)= 400
Number of boys got graduated = (1000 - 400) = 600
Department R:
Total students (Boys + Girls) got graduated = 400 × 2 = 800
Number of girls got graduated = 800 × (3/8)= 300
Number of boys got graduated = (800 - 300) = 500
Department S:
Total students (Boys + Girls) got graduated = 2 × 2 × 200 = 800
Number of girls got graduated = 800 × (1/2) = 400
Number of boys got graduated = (800 - 400) = 400
| Departments | Number of boys graduated | Number of girls graduated |
|---|---|---|
| P | 360 | 240 |
| Q | 600 | 400 |
| R | 500 | 300 |
| S | 400 | 400 |
Solution:
Average number of girls graduated from the all the college = 335
Average number of all the boys graduated from the ll the departments = (360 + 600 + 500 + 400)/4 = 465
Required difference = (465 - 335) = 130
Directions: The following table shows the average number of students (Boys + Girls) got graduation and fractional value of the number of girls who got graduation in four different colleges P, Q, R and S.
| Departments | Average number of (Boys + Girls) got graduation | Fractional value of number of girls got graduation |
|---|---|---|
| P | 1.5x | 2/5 |
| Q | 250% of x | 2/5 |
| R | 400 | 3/8 |
| S | 2x | 1/2 |
Note:
Average number of girls got graduation in all the given colleges = 335
Number of boys who graduated from department T is 50 more than the number of boys who graduated from department Q. Number of girls who graduated from department T is 60 more than the number of girls who graduated from department P. Find the difference between 2/5 of the number of boys who graduated from department T and 2/3 rd of the number of girls who graduated from department T.
85
90
80
60
75
Correct answer
60
Why this answer is correct
General solution:
[2 × 1.5x × (2/5) +2 × 2.5x × (2/5) + 2 × 400 × (3/8) + 2 × 2x × (1/2)] = 1340
[1.2x + 2x + 300 + 2x] = 1340
5.2x = 1040
x = 200
Department P:
Total students (Boys + Girls) got graduated = 1.5 × 2 × 200 = 600
Number of girls got graduated = 600 × (2/5)= 240
Number of boys got graduated = (600 - 240) = 360
Department Q:
Total students (Boys + Girls) got graduated = (250 × x/100) = 2.5 × 2 × 200 = 1000
Number of girls got graduated = 1000 × (2/5)= 400
Number of boys got graduated = (1000 - 400) = 600
Department R:
Total students (Boys + Girls) got graduated = 400 × 2 = 800
Number of girls got graduated = 800 × (3/8)= 300
Number of boys got graduated = (800 - 300) = 500
Department S:
Total students (Boys + Girls) got graduated = 2 × 2 × 200 = 800
Number of girls got graduated = 800 × (1/2) = 400
Number of boys got graduated = (800 - 400) = 400
| Departments | Number of boys graduated | Number of girls graduated |
|---|---|---|
| P | 360 | 240 |
| Q | 600 | 400 |
| R | 500 | 300 |
| S | 400 | 400 |
Solution:
Number of boys graduated from department Q = 600
Number of boys graduated from department T = (600 + 50) = 650
Number of girls graduated from department P = 240
Number of girls graduated from department T = (240 + 60) = 300
Required difference = (650 × 2/5 - 300 × 2/3) = (260 - 200) = 60
The cost price of a cricket bat is 25x and that of one ball is 5x. The marked price of the bat and ball is 20% and 100% more than the cost price, respectively. If the profit percentage on bat and ball is 10% and 50%, respectively, then, find the overall discount percentage?
12%
15.5%
12.5%
16%
None of these
Correct answer
12.5%
Why this answer is correct
Shortcut Trick
Total Marked Price (MP) = Bat MP + Ball MP = (25x × 1.2) + (5x × 2.0) = 30x + 10x = 40x.
Total Selling Price (SP) = Bat SP + Ball SP = (25x × 1.1) + (5x × 1.5) = 27.5x + 7.5x = 35x.
Total Discount = Total MP − Total SP = 40x − 35x = 5x.
Overall Discount % = (Total Discount ÷ Total MP) × 100 = (5x ÷ 40x) × 100 = 12.5%.
∴ The correct answer is 12.5%.
Alternate Method
Given: CP of Bat = 25x, CP of Ball = 5x.
Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = CP × (100 + Profit%) ÷ 100 | Discount% = (MP − SP) ÷ MP × 100.
⇒ Bat MP = 25x + 20% of 25x = 25x + 5x = 30x.
⇒ Ball MP = 5x + 100% of 5x = 5x + 5x = 10x.
⇒ Total MP = 30x + 10x = 40x.
⇒ Bat SP = 25x + 10% of 25x = 25x + 2.5x = 27.5x.
⇒ Ball SP = 5x + 50% of 5x = 5x + 2.5x = 7.5x.
⇒ Total SP = 27.5x + 7.5x = 35x.
⇒ Discount = 40x − 35x = 5x.
⇒ Discount Percentage = (5x ÷ 40x) × 100 = 12.5%.
∴ The correct answer is 12.5%.
Additional Information
Marked Price and Discount
Marked Price is the price displayed on the tag, while Discount is always calculated on the MP to arrive at the Selling Price.
Profit and Cost Price
Profit is the extra amount earned over the Cost Price. Profit % = (SP − CP) ÷ CP × 100.
In 60L of solution of insecticide, the ratio of chemicals and water is 11 : 4. It was decided later insecticide concentration must be further diluted to protect plants and soil. Then how much water should be added to the existing solution so that the ratio of chemical and water be 11: 6?
8 L
6 L
4 L
10 L
12 L
Correct answer
8 L
Why this answer is correct
Shortcut Trick
Initial ratio of chemical to water = 11 : 4
Final ratio of chemical to water = 11 : 6
Since chemical remains constant (11 units), the increase in water = 6 − 4 = 2 units.
Total initial units = 11 + 4 = 15 units → 60 L.
1 unit = 4 L ∴ 2 units = 8 L.
∴ The correct answer is 8 L.
Alternate Method
Given: Total solution = 60 L, Ratio (C:W)initial = 11:4, Ratio (C:W)final = 11:6
Calculations:
⇒ Total initial parts = 11 + 4 = 15
⇒ Quantity of chemical = (11 ÷ 15) × 60 = 44 L
⇒ Quantity of water = (4 ÷ 15) × 60 = 16 L
⇒ Let water added be x L. New water = 16 + x
⇒ 44 ÷ (16 + x) = 11 ÷ 6
⇒ 11 × (16 + x) = 44 × 6
⇒ 16 + x = 4 × 6 = 24
⇒ x = 24 − 16 = 8 L
∴ The correct answer is 8 L.
Additional Information
Mixture Dilution Rule
When adding only one component to a mixture, the quantity of the other component remains constant. Always equate the constant component to find the new total volume.
Alligation Method
For dilution with pure water, the concentration of the second addition is 0% chemical or 100% water. (Q1 ÷ Q2) = (n2 − navg) ÷ (navg − n1).
what approximate value should come in the place of question mark (?) in the following question? (You are not expected to calculate the exact value)
9.87% of [(299.96)2 + (20.24)2] = 999.99% of 4 + ?
8990
9899
9000
9989
N.O.T.
Correct answer
9000
Why this answer is correct
Rules of Approximation:
Concept:
In this type of question, we are expected to calculate approximate value (not exact value), so we can replace the given numbers by their nearest perfect places which makes the calculation easy.
9.87% of [(299.96)2 + (20.24)2] = 999.99% of 4 + ?
Taking approximate value, we get
10% of [(300)2 + (20)2] = 1000% of 4 + ?
Follow BODMAS rule to solve this question, as per the order given below,
Since, we need to find out the approximate value, we can write these values to their nearest integers.
Taking approximate value, we get
10% of [(300)2 + (20)2] = 1000% of 4 + ?
⇒ (10 ÷ 100) × (90000 + 400) = (1000 ÷ 100) × 4 + ?
⇒ (1 ÷ 10) × (90400) = 10 × 4 + ?
⇒ 9040 = 40 + ?
⇒ 9040 -40 = ?
∴ ? = 9000
∴ the approximate value of question mark (?) in the following question is 9000.
What approximate value should come in place of ‘?’ in the following equation:
29.93% of 459.96 + 25.03% of 640.06 + 2.11 = ? × 6.08
50
300
49
30
None of these
Correct answer
50
Why this answer is correct
Solution:
Rules of Approximation:
Concept Used:
Follow BODMAS rule to solve this question, as per the order given below,
Calculation:
30% of 460 + 25% of 640 + 2 = ? × 6
⇒ (30/100) × 460 + (25/100) × 640 + 2 = ? × 6
⇒ 3 × 46 + (640/4) + 2 = ? × 6
⇒ 138 + 160 + 2 = ? × 6
⇒ 300 = ? × 6
⇒ ? = 300/6
∴ ? = 50
What approximate will come in the place of the question mark ‘?’ in the following question?
√1155.91 + √730 + (44.99 ÷ 8.97 - 0.99)0.49 = ?
64
36
49
63
72
Correct answer
63
Why this answer is correct
Solution:
We have to follow the BODMAS rule
Calculation:
√1155.91 + √730 + (44.99 ÷ 8.97 - 0.99)0.49
⇒ √1156 + √730 + (45 ÷ 9 – 1)0.5
⇒ √1156 + √730 + (5 – 1)1/2
⇒ √1156 + √730 + 41/2
⇒ √1156 + √730 + 2
⇒ 34 + 27 + 2
⇒ 63
∴ 63 will come in place of the question mark
What approximate value should come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value)
√3.98 + (√((2.22)2 + 4.85)) × 5.69 = 24.71% of ?
60
70
80
90
23
Correct answer
80
Why this answer is correct
Rules of Approximation:
Concept:
In this type of question, we are expected to calculate approximate value (not exact value), so we can replace the given numbers by their nearest perfect places which makes the calculation easy.
√3.98 + (√((2.22)2 + 4.85)) × 5.69 = 24.71% of ?
Taking approximate value, we get
√4 + (√(22 + 5)) × 6 = 25% of ?
Follow BODMAS rule to solve this question, as per the order given below,
⇒ √4 + (√ (4 + 5)) × 6 = 25% of ?
⇒ √4 + (√9) × 6 = 25% of ?
⇒ 2 + 18 = 25% of ?
⇒ 2 + 18 = 25% of ? (∵ 1.73 ≈ 2)
⇒ 20 = (25/100) × ?
⇒ 20 × 100 / 25 = ?
? = 80
What approximate value should come in place of the question mark (?) in the following question?
292 + 34.12 % of 700 ÷ ∛2744.21 = ? + ∛2197
945
855
895
845
865
Correct answer
845
Why this answer is correct
Rules of Approximation:
Concept:
In this type of question, we are expected to calculate approximate value (not exact value), so we can replace the given numbers by their nearest perfect places which makes the calculation easy.
292 + 34.12 % of 700 ÷ ∛2744.21 = ? + ∛2197
Taking approximate value, we get
292 + 34 % of 700 ÷ ∛2744 = ? + ∛2197
Follow BODMAS rule to solve this question, as per the order given below,
⇒ 841 + (34 ÷ 100) × 700 ÷ 14 = ? + 13
⇒ 841 + (238 ÷ 14) = ? + 13
⇒ 841 + 17 = ? + 13
⇒ 858 = ? + 13
⇒ ? = 858 – 13
⇒ ? = 845
∴ The value of ? is 845
Ram, Shyam, and Rohan working together then they can complete the work in 90 days. Find in how many days Rohan and Shyam can complete the 44.44% of work if the efficiency of ram is 20% more than Shyam and efficiency of Rohan is 16.66% less than ram.
75
16
64
67
None
Correct answer
64
Why this answer is correct
Shortcut Trick
Efficiency ratio Ram : Shyam = 6 : 5 (since 20% = 1/5)
Efficiency ratio Rohan : Ram = 5 : 6 (since 16.66% = 1/6)
Combined Efficiency Ram : Shyam : Rohan = 6 : 5 : 5
Total Work = 90 × (6 + 5 + 5) = 1440 units
Target Work = 44.44% of 1440 = 4/9 × 1440 = 640 units
Required Days = 640 ÷ (5 + 5) = 64 days
∴ The correct answer is 64 days.
Alternate Method
| Person | Efficiency Ratio | Normalized Efficiency |
|---|---|---|
| Ram | 120% of Shyam | 6x |
| Shyam | Base | 5x |
| Rohan | (1 − 1/6) of Ram | 5x |
Given: Time taken by (Ram + Shyam + Rohan) = 90 days.
Calculations:
⇒ Total Efficiency = 6x + 5x + 5x = 16x
⇒ Total Work = Efficiency × Time = 16x × 90 = 1440x
⇒ Required Work = 44.44% of Total Work = (4/9) × 1440x = 640x
⇒ Efficiency of (Shyam + Rohan) = 5x + 5x = 10x
⇒ Required Days = 640x ÷ 10x = 64 days
∴ The correct answer is 64 days.
Additional Information
Percentage to Fraction Conversion
Significant values used: 16.66% = 1/6 and 44.44% = 4/9. Converting percentages to fractions simplifies ratio calculations.
Work, Time, and Efficiency Relationship
Total Work = (Combined Efficiency of workers) × (Total Time taken). Efficiency is inversely proportional to time for a constant work.
Ratio Balancing
To find a combined ratio like A:B:C, ensure the common term has the same value in both individual ratios (A:B and B:C or A:C).
Solving previous year questions by topic helps identify recurring concepts, question patterns, and frequently asked sub-topics. Attempt each question, review the step-by-step explanation, and note down key formulas or facts for quick revision before the exam.