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 19** 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 19** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. 9900

B. 9750

C. 9450

D. 9000

**Answer**: Option A

**Explanation**:

Largest 4-digit number is 9999.

Remember: The question may be asked in a tricky way. Here, the largest number does not mean H.C.F.. We have to find a number which is divisible by 12, 15 and 18

The required largest number must be divisible by the L.C.M. of 12, 15 and 18

L.C.M. of 12, 15 and 18

12 = 2 × 2 × 3

15 =5 × 3

18 = 2 × 3 × 3

L.C.M. = 180

Now divide 9999 by 180, we get remainder as 99

The required largest number = (9999 – 99) =9900

Number 9900 is exactly divisible by 180

**2. 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 × 13 = 195

11y = 11 × 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 × 2145 = 27885

Product of two numbers = 195 × 143 = 27885

Hence, the calculated answer is correct.

**3. 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 × 37y = 4107

xy = 3

3 is the product of (1 and 3)

x = 1 and y = 3

37x = 37 × 1 =37

37y = 37 × 3 = 111

Greater number = 111

**4. Find the sum of two numbers, which are greater than 29 and have H.C.F. and L.C.M. of 29 and 4147 respectively. **

- A. 858

B. 696

C. 1050

D. 4147

**Answer**: Option B

**Explanation**:

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

Product of two numbers = 29 × 4147 = 120263

Given: Two numbers are greater than 29.

Therefore, let the two numbers be 29x and 29y.

So, 29x × 29y = 120263

Xy = 143

Factors of 143 are: 1, 11, 13, and 143

Case: 1) If we consider factors of 143 as 1 and 143 (co-primes), then we get the value of two numbers x and y = (29 and 4147) —— (Which is wrong: As it is given that, the numbers are greater than 29)

Case: 2) If we consider factors of 143 as 11 and 13 (co-primes), then we get the value of two numbers x and y = (319, 377) —— (These two values are greater than 2. So, it is the correct answer)

Therefore, the two numbers are 319 and 377.

Sum of two numbers = 319 + 377 = 696

**5. Find the greatest number, which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.
**

- A. 5 : 28 : 00 hrs

B. 5 : 30: 00 hrs

C. 5: 38 : 00 hrs

D. 5 : 40: 00 hrs

**Answer**: Option A

**Explanation**:

The number on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

Hence, make the dividend completely divisible by the divisor. This is possible if we subtract the remainder from the dividend.

Therefore,

1657 – 6 = 1651

2037 – 5 = 2032

H.C.F. of 1651 and 2032 is 127. 127 is the common factor.

127 × 13 = 1651

Thus by adding 6, we get 1651 + 6 = 1657

127 is the correct answer.

- A. 5 : 28 : 00 hrs

B. 5 : 30 : 00 hrs

C. 5 : 38 : 00 hrs

D. 5 : 40 : 00 hrs

**Answer**: Option B

**Explanation**:

Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.

Therefore, find the L.C.M. of 40, 72 and 108.

L.C.M. of 40, 72 and 108 = 1080

The traffic lights will change again after 1080 seconds = 18 min

The next simultaneous change takes place at 5 : 38 : 00 hrs.

**2. A rectangular courtyard 4.55 meters long and 5.25 meters wide is paved exactly with square tiles of the same size. Find the largest size of the tile used for this purpose? **

- A. 25 cm

B. 21 cm

C. 45 cm

D. 35 cm

**Answer**: Option D

**Explanation**:

Here, we are asked to find the largest size of tile. Therefore, calculate H.C.F.

Step 1: Covert numbers without decimal places i.e 455 cm and 525 cm

Step 2: Find the H.C.F. of 455 and 525

H.C.F. of 455 and 525 = 35 cm

Hence, the largest size of the tile is 35 cm.

**3. John, Smith, and Kate start at the same time, the same point and in the same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point? **

- A. 30 min

B. 25 min

C. 20 min

D. 15 min

**Answer**: Option B

**Explanation**:

L.C.M. of 250, 300 and 150 = 1500 sec

Dividing 1500 by 60 we get 25, which mean 25 minutes.

John, Smith and Kate meet after 25 minutes.

**4. If Suresh distributes his pens in the ratio of \(\frac{1}{2 }\) : \(\frac{1}{4 }\) : \(\frac{1}{5 }\) : \(\frac{1}{7 }\) between his four friends A, B, C and D, then find the total number of pens Suresh should have?**

- A. 153

B. 150

C. 100

D. 125

**Answer**: Option A

**Explanation**:

Here, A : B : C : D = \(\frac{1}{2 }\) :\(\frac{1}{4 }\) : \(\frac{1}{5 }\) : \(\frac{1}{7 }\)

1) L.C.M of 2, 4, 5, 7 is 140

2) Find the number of pens each friend received ——— (To find no. of pens each friend has, multiply the ratio with the L.C.M. calculated)

A = (\(\frac{1}{2 }\)) x 140 = 70

B = (\(\frac{1}{4 }\)) x 140 = 35

C = (\(\frac{1}{5 }\)) x 140 = 28

D = (\(\frac{1}{7 }\)) x 140 = 20

3) Total number of pens = (70 x + 35 x + 28 x + 20 x) = 153 x

Minimum number of pens (x) = 1

Therefore, total number of pens = 153 pens.

**5. A man, his wife, and daughter worked in a garden. The man worked for 3 days, his wife for 2 days and daughter for 4 days. The ratio of daily wages for man to women is 5 : 4 and the ratio for man to daughter is 5 : 3. If their total earnings are mounted to Rs. 105, then find the daily wage of the daughter. **

- A. Rs. 15

B. Rs. 12

C. Rs. 10

D.

**Answer**: Option D

**Explanation**:

Assume that the daily wages of man, women and daughter are Rs 5x, Rs. 4x, Rs 3x respectively.

Multiply (no. of days) with (assumed daily wage) of each person to calculate the value of x.

[3 x (5x)] + [2 x (4x)] + [4 x (3x)] = 105

[15x + 8x + 12x] = 105

35x = 105

x = 3

Hence, man’s daily wage = 5x = 5 x 3 = Rs. 15

Wife’s daily wage = 4x = 4 x 3 = Rs. 12

Daughter’s daily wage = 3x = 3 x 3 = Rs. 9

- A. Rs. 15.25

B. Rs. 17

C. Rs. 12

D. Rs. 5

** Answer**: Option D

**Explanation**:

Toatal bill paid by Amit, Raju and Ram = ( 50 + 55 +75 ) = Rs. 180

Let amount paid by Amit, Raju and Ram be Rs. 3x, 4x and 5x respectively.

Therefore, (3x + 4x + 5x ) = 180

12x = 180

x = 15

Therefore, the amount paid by,

Amit = Rs. 45

Raju = Rs. 60

Ram = Rs. 75

But actually as given in the question, Amit pays Rs. 50, Raju pays Rs. 55 and Ram pays Rs. 80. Hence, Amit pays Rs. 5 less than the actual amount to be paid. Hence he needs to pay Rs. 5 to Raju settle the amount.

**2. If the length is increased by 25%, by what percent the width of a rectangle should be decreased so as to keep the area same.**

- A. 25%

B. 20%

C. 30%

D. 10%

** Answer**: Option B

**Explanation**:

Let the original length be l and the width be b

Therefore, the area = l*b

Now, as the length is increased by 25%, the new length is (1.25*l) and let the new width be x.

As the area is same, 1.25*l*x = l*b

x = b/1.25 = 0.8b

Therefore, the width is to be decreased by 20%.

**3. A bag contains an equal number of 25 paise, 50 paise, and one rupee coins respectively. If the total value is Rs 105, how many types of each type are present?**

- A. 75 coins

B. 60 coins

C. 30 coins

D. 25 coins

** Answer**: Option B

**Explanation**:

Bag consists of 25 paise, 50 paise and 1 rupee (100 paise) so the ratio becomes 25 : 50 : 100 or 1 : 2 : 4

Total value of 25 paise coins = (\(\frac{1}{7 }\) ) x 105 = 15

Total value of 50 paise coins = (\(\frac{2}{7 }\)) x 105 = 30

Total value of 100 paise coins = (\(\frac{4}{7 }\)) x 105 = 60

No. of 25 paise coins = 15 x 4 = 60 coins

No. of 50 paise coins = 30 x 2 = 60 coins

No. of 1 rupee coins = 60 x 1 = 60 coins

**4. A purse contains 342 coins consisting of one rupees, 50 paise and 25 paise coins. If their values are in the ratio of 11 : 9 : 5 then find the number of 50 paise coins?**

- A. 180

B. 150

C. 162

D. 99

** Answer**: Option C

**Explanation**:

Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively.

No. of 1 rupee coins = \(\frac{11x }{1 }\) = 11x

No. of 50 paise coins = \(\frac{9x }{0.5 }\) = 18x

No. of 25 paise coins = \(\frac{15x }{0.25 }\) = 20x

11x + 18x + 9x = 342

38x = 342

x = 9

Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins

No. of 50 paise coins = 18 x 9 = 162 coins

No. of 25 paise coins = 20 x 9 = 180 coins

**5. When folded into two equal halves a rectangular sheet had a perimeter of 48cm for each part folded along one set of sides and the same is 66cm when folded along the other set of sides. Find the area of the sheet.**

- 1584

B. 1120

C. 792

D. 1320

** Answer**: Option B

**Explanation**:

Let the sheet be folded along its breadth and its perimeter = 48cm

Therefore, (\(\frac{l }{2 }\) + b) = 48 …… (i)

Now, let the sheet be folded along its length, and the perimeter = 66cm

(l + \(\frac{b }{2 }\))= 66 …… (ii)

Solving (i) and (ii), we get,

l = 56cm, b = 20cm

Area = l*b

Area = 1120 cm2