 # QUANT QUIZ FOR SSC EXAMS , RAILWAYS GROUP D & NTPC 2019 : Part – 46 ## QUANT QUIZ FOR SSC EXAMS , Railways GROUP D & NTPC 2019

Q1. If a commission 0f 10% is given on mark price of Truck the dealer gains 20% if the commission is increase to 15% the gain % is?

(a) 40/3

(b) 10

(c) 20

(d) 15

(e) None of these

Q2. The CP of two dozen mangoes is Rs 32, after selling 18 mangoes at 12 Rs per dozen, the shopkeeper reduced the rate as Rs 4 per dozen. Then find the loss percentage?

(a) 15

(b) 20

(c) 25

(d) 37.5

(e) None of these

Q3. A can complete a task in 15 days B is 50% more efficient than A. Both A and B started working together on the task and after few days B left task and A finished the remaining 13 of the given work. For how many days A and B worked together.

(a) 3

(b) 5

(c) 4

(d) 6

(e) 2

Q4. A boat can travel 9.6 km downstream in 36 min. If speed of the water current is 10% of the speed of the boat in downstream, how much time will boat take to travel 19.2 km upstream?

(a) 2 hours

(b) 3 hours

(c) 1.25 hours

(d) 1.5 hours

(e) 1 hour

Q5. A started a business with a initial investment of Rs. 1200. ‘X’ month after the start of business, B joined A with on initial investment of Rs. 1500. If total profit was 1950 at the end of year and B’s share of profit was 750. Find ‘X’.

(a) 5 month

(b) 6 month

(c) 7 month

(d) 8 month

(e) 9 month

Q6. Ratio between curved surface area and total surface area of a circular cylinder is 3: 5. If curved surface area is 1848 cm3 then what is the height of cylinder?

(a) 28

(b) 14

(c) 17

(d) 21

(e) 7

Q7. Train A completely crosses train B which is 205 m long in 16 second. If they are travelling in opposite direction and sum of speed of both are 25 m/s. then find the difference (in meter) between lengths of both trains.

(a) 5

(b) 6

(c) 8

(d) 10

(e) 12

Q8. Present age of A is 3 years less than present age of B. Ratio of B’s age 5 year ago and A’s age 4 year hence is 3 : 4 then find present age (in years) of A.

(a) 20

(b) 17

(c) 23

(d) 26

(e) 29

Q9. Elena’s age after 15 years will be 5 times of her age 5 years back. What is the present age of Elena?

(a) 10

(b) 37

(c) 35

(d) 11

(e) None of these

Q10. A mess has provisions for 60 days. If after 15 days 500 men strengthen then and the food lasts 40 days longer, how many men are there in the mess?

(a) 3500

(b) 4000

(c) 6000

(d) 8000

(e) None of these

Directions (11-15): In each of the following questions two equations are given followed by 5 options. You have to solve the equations and give answer:

Q.11 I. 5x² + 29x + 20 = 0
II. 25y² + 25y + 6 = 0
(a) if x < y
(b) if x ≤ y
(c) relationship between x and y cannot be determined
(d) if x ≥ y
(e) if x > y

Q.12 I. 3x² – 16x + 21 = 0
II. 3y² – 28y + 65 = 0
(a) if x < y
(b) if x ≤ y
(c) relationship between x and y cannot be determined
(d) if x ≥ y
(e) if x > y

1. 13 I. x² + x – 12 = 0

II. y² + 2y – 8 = 0

(a) if x < y
(b) if x ≤ y
(c) relationship between x and y cannot be determined
(d) if x ≥ y
(e) if x > y

1. 14 I. 4x² – 13x + 9 = 0

II. 3y² – 14y + 16 = 0

(a) if x < y
(b) if x ≤ y
(c) relationship between x and y cannot be determined
(d) if x ≥ y
(e) if x > y

Q.15 I. 16x² + 20x + 6 = 0
II. 10y² + 38y + 24 = 0

(a) if x < y
(b) if x ≤ y
(c) relationship between x and y cannot be determined
(d) if x ≥ y
(e) if x > y

## Solutions:

Sol 1. (a) Let S.P. = Rs. 100

After commission, price = Rs. 90

CP = 100/120×90 = Rs. 75

Now, commission = 15%

Gain % = 85−75/75×100

=40/3%

Sol 2. (d) Total CP = Rs. 32

Total SP = 12 + 6 + 2 = Rs. 20

Loss % =12/32×100

= 37.5%

Sol 3. (c) A can complete work in 15 days.

And B will complete work in 10 days.

They together will complete whole work in =15×10/25=6 days

A and B together worked for = 6 × 2/3= 4 days

Sol 4. (d) Speed of downstream=9.6/36kmmin=16 km/hr

Speed of current = 1.6 km/hr

Let speed of man in still water = 𝑥

So, 𝑥 = 16 – 1.6 = 14.4 km/hr

Required time in upstream=19.2/14.4−1.6

=1.5 hours

Sol 5. (b) Ratio of profit of A and B = 1200:750 = 8:5

So, 1200×12/1500×𝑦 = 8/5

𝑦=6 months

x= 6 month

Sol 6. (d) 2πrh/2πr(r+h) = 3/5

2h = 3r

And 2πrh = 1848

2×22/7×2/3h×h=1848

h=21

Sol 7. (d) In 16 second distance covered by both A and B = 16 × 25 = 400 m

So length of A = 400 – 205 = 195

Required difference = 10 m

Sol 8. (a) Let B’s age = 𝑥

So A’s age = 𝑥−3

𝑥−5/𝑥+1=3/4

𝑥=23

A’s age = 23−3=20 𝑦𝑒𝑎𝑟𝑠

Sol 9. (a) Let Elena’s age = x

x + 15 = 5 (x – 5)

x = 10 years

Sol 10. (b) Let No. of soldiers = x

60 × x = 15x + 40 (x + 500)

60x =15x + 40x + 20000

5x = 20000

x = 4000

S11. Ans. (a)
Sol.
I. 5x² + 29 + 20 = 0
⇒ 5x² + 25x + 4x + 20 = 0
⇒ (x + 5) (5x + 4) = 0
⇒ x = –5, –4/5

1. 25y² + 25y + 6 = 0
⇒ 25y² + 15y + 10y + 6 = 0
⇒ (5y + 3) (5y + 2) = 0
⇒ y = – 3/5, –2/5
y > x

S12. Ans.(a)
Sol.
I. 3x² – 16x + 21 = 0
⇒ 3x² – 9x – 7x + 21 = 0
⇒ (x – 3) (3x – 7) = 0
⇒ x = 3, 7/3

II. 3y² – 28y + 65 = 0
⇒ 3y² – 15y – 13y + 65 = 0
⇒ (y – 5)(3y – 13) = 0
⇒ y = 5, 13/3

y > x

S13. Ans.(c)
Sol.
I. x² + x – 12 = 0
⇒ x² + 4x – 3x – 12 = 0
⇒ (x + 4) (x – 3) = 0
⇒ x = 3, –4

II. y² + 2y – 8 = 0
⇒ y² + 4y – 2y – 8 = 0
⇒ (y + 4) (y – 2) = 0
⇒ y = – 4, 2
No relation

S14. Ans.(c)
Sol.
I. 4x² – 13x + 9 = 0
⇒ 4x² – 4x – 9x + 9 = 0
⇒ (x – 1) (4x – 9) = 0
⇒ x = 1, 9/4

II. 3y² – 14y + 16 = 0
⇒ 3y² – 6y – 8y + 16 = 0
⇒ (y – 2) (3y – 8) = 0
⇒ y = 2, 8/3
No relation

S15. Ans.(e)
Sol.
I. 16x² + 20x + 6 = 0
⇒ 8x² + 10x + 3 = 0
⇒ 8x² + 4x + 6x + 3 = 0
⇒ (2x + 1) (4x + 3) = 0
⇒ x = –1/2, –3/4

1. 10y² + 38y + 24 = 0

⇒ 5y² + 19y + 12 = 0

⇒ 5y² + 15y + 4y + 12 = 0

⇒ (y + 3) (5y + 4) = 0

y = –3, –4/5

x > y

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May 23, 2019

## Free Reasoning Quiz for SBI , IBPS Clerk & PO – PART – 42 