MATHS : QUIZ for SSC

ANSWERS WITH SOLUTION:
1(B)Explanation:
S.I. on Rs. 1600 = T.D. on Rs. 1680.
 Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%.
 Time = ((100 x 80)/(1600 x 15)year = 1/3 year = 4 months.

2( C)Explanation:
B.D. for  3/2 years =  Rs. 558.
B.D. for 2 years = (558 x 2/3 x 2)
                                      = Rs. 744
T.D. for 2 years = Rs. 600.
 Sum = (B.D. x T.D)/(B.D. – T.D) = Rs. (744 x 600/144) = Rs. 3100.
Thus, Rs. 744 is S.I. on Rs. 3100 for 2 years.
Rate = (100 x 744)/(3100 x 2)%= 12%

3(D)

4( A)Explanation :F = Rs. 3000
R = 10%
Date on which the bill is drawn = 14th July at 5 months
Nominally Due Date = 14th December
Legally Due Date = 14th December + 3 days = 17th December
Date on which the bill is discounted = 5th October
Unexpired Time
= [6th to 31st of October] + [30 Days in November] + [1st to 17th of December]
= 26 + 30 + 17 = 73 Days
=73/365 year = 1/5 year
BD = Simple Interest on the face value of the bill for unexpired time =FTR/100
=3000×1/5×10/100 =30×1/5×10 = Rs. 60

5( A)Explanation :
Let the investment in scheme A be Rs.x
and the investment in scheme B be Rs.(13900 – x)
We know that SI = PRT/100
Simple Interest for Rs.x in 2 years at 14% p.a. = x×14×2/100=28x/100
Simple Interest for Rs.(13900 – x) in 2 years at 11% p.a
. =(13900-x)×11×2/100=22(13900-x)/100
Total interest = Rs.3508
28x/100+22(13900-x)/100=3508
28x+305800-22x=350800
6x=45000
x=450006=7500
Investment in scheme B = 13900 – 7500 = Rs.6400

6(D)Explanation :
Let the sum of money be x
then
x×4×8/100=560×12×8/100
x×4×8=560×12×8
x×4=560×12
x=560×3=1680

7(D)Explanation :
Given that simple interest for 2 years is Rs.800
i.e., Simple interest for 1st year is Rs.400
and simple interest for 2nd year is also Rs.400
Compound interest for 1st year will be 400
and Compound interest for 2nd year will be 832 – 400 = 432
you can see that compound interest for 2nd year is more than simple interest for 2nd year by 432 – 400 = Rs.32
i.e, Rs. 32 is the interest obtained for Rs.400 for 1 year
Rate, R = 100×SI/PT= (100×32)/(400×1)= 8%
Difference between compound and simple interest for the 3rd year
= Simple Interest obtained for Rs.832
=PRT/100=832×8×1/100= Rs. 66.56
Total difference between the compound and simple interest for 3 years
= 32 + 66.56 = Rs.98.56

8(A)Explanation :
Let the investment in scheme A be Rs.x
Then the investment in scheme = Rs.(15000-x)
Compound interest for Rs.x = P(1+R/100)^T-P= x(1+5/100)^2-x
Compound interest for Rs. (15000-x) = P(1+R/100)^T-P=
(15000-x)(1+10/100)^2-(15000-x)
Total compound interest = x(1+5/100)^2-x+(15000-x)(1+10/100)^2-(15000-x)
=x(1+10/100+25/10000)-x+(15000-x)(1+20/100+100/10000)-(15000-x)
=x(10/100+25/10000)+(15000-x)(20/100+100/10000)
=1025x/10000+(15000-x)((2100/10000)
=1025x/10000+15000×2100/10000-2100x/10000
=-1075x/10000+3150
Given that total compound interest = Rs. 2075
-1075x/10000+3150=2075
=>1075x/10000=1075
=>x/10000=1
=>x=10000
i.e, Amount invested in Scheme A = Rs. 10000

9(C)Explanation :
cost of x metres of wire = Rs. d
cost of 1 metre of wire = Rs.(d/x)
cost of y metre of wire = Rs.(y×d/x)=Rs. (yd/x)

10(D)Explanation :
Let the number of revolutions made by the larger wheel be x
More cogs, less revolutions (Indirect proportion)
Hence we can write as
Cogs 6 : 14} :: x : 21
 6×21= 14×x
6×3 = 2×x
3×3 = x
x=9