Quantitative Aptitude - Arithmetic Ability Questions

Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'

=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.

=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15

Therefore, 15 sweets were each student originally supposed to get.

Let the two no's be a and b;
Given product of the no's is p = ab;
If the each nos is increased by 2 then the new product will be
(a+2)(b+2) = ab + 2a + 2b + 4
= ab + 2(a+b) + 4
= p + 2(a+b) + 4
Hence the new product is (p+4) times greater than twice the sum of the two original numbers.

Distance = 70 x 1 ½ = 105 km

Relative Speed = 85 – 70 = 15

Time = 105/15 = 7 hrs

4:30 + 7 hrs = 11.30 p.m.

Let the time required by x seconds.

Then, More cloth means More time (Direct Proportion)

So, 0.128 : 1 :: 25 : x

⇒x=1×250.128 => x = 195.31

 So time will be approx 195 seconds

We know that the day repeats every 7 days, 14 days, 21 days,...

So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday.

Hence, after 2 weeks i.e, 14 days the day repeats and is the same day. 

Let varun's age = x yrs.

From the given conditions,

x+1  =  2(x-12)

x = 25 years.

Length of the first carpet = (1.44)(6) = 8.64 cm

Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)

= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m

Cost of the second carpet = (45)(12.96 x 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40

Let p = Rs. 100.

Then, S.I is Rs. 50 and time = 5 years. 

∴ R = 100×50100×5= 10% p.a.

Now, p = Rs. 12,000 , T = 3 years and R = 10% p.a.

C.I. = Rs. 12000×1+101003-1= Rs. 3972