A candidate with **quantitative aptitude** knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an **important** measure for a prospective business executive’s **abilities**.

The article **SSC CPO Quantitative Aptitude Quiz 21** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc. The article **SSC CPO Quantitative Aptitude Quiz 21** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. 6.2 years

B. 7.7 years

C. 8.7 years

D. 10 years

**Answer**: Option C

**Explanation**:

1) Let Johnâ€™s present age be x

2) Johnâ€™s age before 5 years = (x – 5)

3) Johnâ€™s age after 10 years = (x + 10)

We are given that, Johnâ€™s age after 10 years (x + 10) is 5 times his age 5 years back (x â€“ 5)

Therefore,

(x + 10) = 5 (x â€“ 5)

Solving the equation, we get

x + 10 = 5x â€“ 25

4x = 35

x = 8.75 years

**2. Rahul is 15 years elder than Rohan. If 5 years ago, Rahul was 3 times as old as Rohan, then find Rahul’s present age.**

- A. 32.5 years

B. 27.5 years

C. 25 years

D. 24.9 years

**Answer**: Option B

**Explanation**:

1) Let age of Rohan be y

2) Rahul is 15 years elder than Rohan = (y + 15). So Rahulâ€™s age 5 years ago = (y + 15 â€“ 5)

3) Rohanâ€™s age before 5 years = (y â€“ 5)

5 years ago, Rahul is 3 times as old as Rohan

(y + 15 â€“ 5) = 3 (y â€“ 5)

(y + 10) = (3y â€“ 15)

2y = 25

y = 12.5

Rohanâ€™s age = 12.5 years

Rahulâ€™s age = (y + 15) = (12.5 + 15) = 27.5 years

**3. One year ago, ratio of Harry and Peter ageâ€™s was 5 : 6 respectively. After 4 years, this ratio becomes 6 : 7. How old is Peter? **

- A. 25 years

B. 26 years

C. 31 years

D. 35 years

**Answer**: Option C

**Explanation**:

Hint: If ages in the numerical are mentioned in ratio A : B, then A : B will be Ax and Bx.

We are given that age ratio of Harry : Pitter = 5 : 6

1) Harryâ€™s age = 5x and Peterâ€™s age = 6x

2) One year ago, their age was 5x and 6x. Hence at present, Harryâ€™s age = 5x +1 and Peterâ€™s age = 6x +1

3) After 4 years,

Harryâ€™s age = (5x +1) + 4 = (5x + 5)

Peterâ€™s age = (6x +1) + 4 = (6x + 5)

4) After 4 years, this ratio becomes 6 : 7. Therefore,

\(\frac{Harryâ€™s Age}{6}\) = \(\frac{Peterâ€™s Age}{7}\)

\(\frac{(5x + 5) }{(6x + 5)}\) = \(\frac{6 }{7}\)

7 (5x + 5) = 6 (6x + 5)

X = 5

Peterâ€™s present age = (6x + 1) = (6 x 5 + 1) = 31 years

Harryâ€™s present age = (5x + 1) = (5 x 5 + 1) = 26 years

**4. The age of mother 10 years ago was 3 times the age of her son. After 10 years, the motherâ€™s age will be twice that of his son. Find the ratio of their present ages. **

- A. 11 : 7

B. 9 : 5

C. 7 : 4

D. 7 : 3

**Answer**: Option D

**Explanation**:

We are given that, age of mother 10 years ago was 3 times the age of her son

So, let age of son be x and as motherâ€™s age is 3 times the age of her son, let it be 3x, three years ago.

At present: Motherâ€™s age will be (3x + 10) and sonâ€™s age will be (x + 10)

After 10 years: Motherâ€™s age will be (3x + 10) +10 and sonâ€™s age will be (x + 10) + 10

Motherâ€™s age is twice that of son

(3x + 10) +10 = 2 [(x + 10) + 10]

(3x + 20) = 2[x + 20]

Solving the equation, we get x = 20

We are asked to find the present ratio.

(3x + 10) : (x + 10) = 70 : 30 = 7 : 3

**5. Sharad is 60 years old and Santosh is 80 years old. How many years ago was the ratio of their ages 4 : 6?
**

- A. 10 years

B. 15 years

C. 20 years

D. 25 years

**Answer**: Option C

**Explanation**:

Here, we have to calculate: How many years ago the ratio of their ages was 4 : 6

Let us assume x years ago

At present: Sharad is 60 years and Santosh is 80 years

x years ago: Sharadâ€™s age = (60 â€“ x) and Santoshâ€™s age = (80 â€“ x)

Ratio of their ages x years ago was 4 : 6

\(\frac{(60 â€“ x) }{(80 â€“ x)}\) = \(\frac{4 }{6}\)

6(60 â€“ x) = 4(80 â€“ x)

360 â€“ 6x = 320 â€“ 4x

x = 20

Therefore, 20 years ago, the ratio of their ages was 4 : 6

- A. 1960

B. 2080

C. 2024

D. 2100

**Answer**: Option D

**Explanation**:

The two conditions that decide that a year is a leap year or not is:

â€¢ For a year to be a leap year, it should be divisible by 4.

â€¢ No century is a leap year unless it is divisible by 400.

Hence, the year 2100 is not a leap year as it is not divisible by 400.

**2. The last day of the century cannot be:**

- A. Sunday

B. Wednesday

C. Friday

D. Saturday

**Answer**: Option D

**Explanation**:

100 years have 5 odd days. Hence the last day of the 1st century is a Friday.

200 years have 10 odd days or 1 week + 3 odd days. Hence, the last day of the 2nd century is a Wednesday.

300 years have 15 odd days or 2 weeks + 1 odd day. Hence, the last day of the 3rd century is a Monday.

400 years have 0 odd days. Hence, the last day of the 4th century is a Sunday.

**3. 5 times a positive number is less than its square by 24. What is the integer?**

- A. 5

B. 8

C. 7.5

D. 9

**Answer**: Option B

**Explanation**:

Let the unknown number be x.

5 times a positive number = 5x

5 times a positive number is less than its square by 24

\({x }^{2}\) â€“ 5x = 24

\({x }^{2}\) + 3x â€“ 8x â€“ 24

x (x + 3) â€“ 8(x + 3)

(x – 8) (x + 3)

x = 8

8 is the required integer.

**4. H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers. **

- A. 230, 140

B. 215, 130

C. 195, 143

D. 155, 115

**Answer**: Option C

**Explanation**:

Hint: Product of two numbers = Product of their H.C.F. and L.C.M.

Given:

1) H.C.F. of two numbers = 13

2) The numbers are in the ratio of 15: 11

Let the two numbers be 15y and 11y

H.C.F. is the product of common factors

Therefore, H.C.F. is y. So y = 13

The two numbers are:

15y = 15 Ã— 13 = 195

11y = 11 Ã— 13 = 143

We can cross-check the answer using the trick. (Product of two numbers = Product of their H.C.F. and L.C.M.)

Product of H.C.F. and L.C.M. = 13 Ã— 2145 = 27885

Product of two numbers = 195 Ã— 143 = 27885

Hence, the calculated answer is correct.

**5. If the product and H.C.F. of two numbers are 4107 and 37 respectively, then find the greater number. **

- A. 111

B. 222

C. 332

D. 452

**Answer**: Option A

**Explanation**:

4107 is the square of 37.

So let two numbers be 37x and 37y.

37x Ã— 37y = 4107

xy = 3

3 is the product of (1 and 3)

x = 1 and y = 3

37x = 37 Ã— 1 =37

37y = 37 Ã— 3 = 111

Greater number = 111

OR

Hint:

Product of two numbers = Product of their H.C.F. and L.C.M.

Product of two numbers = 4107

4107 = 37 Ã— L.C.M

L.C.M. = \(\frac{4107 }{37}\) = 111

The greater number = 111

- A. 200 sec

B. 190 sec

C. 210 sec

D. 150 sec

** Answer**: Option B

**Explanation**:

Time taken by B run 1000 meters = \(\frac{(1000 * 10)}{50 }\)= 200 sec.

Time taken by A = 200 – 10 = 190 sec.

**2. A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?**

- A. 111.12 m

B. 888.88 m

C. 777.52 m

D. 756.34 m

** Answer**: Option A

**Explanation**:

A runs 1000 m while B runs 900 m and C runs 800 m.

The number of meters that C runs when B runs 1000 m,

= \(\frac{(1000 * 800)}{9 }\) = \(\frac{8000}{9 }\) = 888.88 m.

B can give C = 1000 – 888.88 = 111.12 m.

**3. In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?**

- A. 57.5 m

B. 127.5 m

C. 150.7 m

D. 98.6 m

** Answer**: Option B

**Explanation**:

When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.

When B runs 900 m, distance that C runs = \(\frac{(900 * 700) }{800 }\) = \(\frac{6300 }{8 }\) = 787.5 m.

In a race of 1000 m, A beats C by (1000 – 787.5) = 212.5 m to C.

In a race of 600 m, the number of meters by which A beats C = \(\frac{(600 * 212.5) }{1000 }\) = 127.5 m.

**4. In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?**

- A. 50

B. 40

C. 25

D. 15

** Answer**: Option C

**Explanation**:

A scores 60 while B score 40 and C scores 30.

The number of points that C scores when B scores 100 = \(\frac{(100 * 30) }{40 }\) = 25 * 3 = 75.

In a game of 100 points, B gives (100 – 75) = 25 points to C.

**5. What sum of money will produce Rs.70 as simple interest in 4 years at 3 \(\frac{1 }{2}\) percent?**

- 1584

B. 1120

C. 792

D. 1320

** Answer**: Option B

**Explanation**:

70 = \(\frac{P*4*7 }{2 }\) X \(\frac{1 }{100 }\)

P = 500