Solutions S1. Ans.(c) Sol. Try Using option and divide 64329 With given option S2. Ans.

32 Pages·2017·0.95 MB·English

Solutions

S1. Ans.(c)
Sol.
Try Using option and divide 64329
With given option

S2. Ans.(a)
Sol. Let the fraction be x/y.
According to the questions,
x 4 4 15
of + =
y 7 7 14
x 4 4 15
× + =
y 7 7 14
x 4 1
× =
y 7 2
x 7
=
y 8

S3. Ans.(b)
Sol. Let the no. of boys = x
and no. of girls = y
According to the questions,

x + y = 14 …(ii)
Solving equation (i) & equation (ii)
x = 8 & y = 6
Total = 14

S4. Ans.(d)
Sol. Let the numbers be LCM (14 and 4) = 28
According to the question, difference
3 3
×28− ×28
4 4
= 21 – 6
= 15
This 15r → 150
1r → 10
Thus, the given number = 280

S5. Ans.(d)
Sol. The sum of first 50 odd natural number = (50)² = 2500

S6. Ans.(d)
Sol. Let the number be = 1000x + 100y + 10x + y
= 10x(100 + 1) + y(100 + 1)
= (101) (10x + y)
Thus, the number is divisible by 101.

S7. Ans.(b)
Sol. Since, the difference is same. So, the number whose product is greater will be smaller &
vice-versa.
Here, largest is √5−√3.

S8. Ans.(a)
Sol. Let the 3rd no. be 200.

First no. is 90% of the second number.

S9. Ans.(d)
Sol. x² + (x + 1)² + (x + 2)² = 110
3x² + 6x = 105
x² + 2x – 35 = 0
(x + 7) (x - 5) = 0
x = -7, 5
Smallest number = 5

S10. Ans.(b)
Sol. Let the maximum marks be x.
20 42
x+10 = x−1
100 100
22x
11 =
100
22% → 11
1
1% →
2
100% → 50
Thus, the maximum marks = 50

S11. Ans.(b)
Sol.
E : S
5 : 3

New saving = 896 – 575 = 321
% increment = 7%

S12. Ans.(c)
Sol. According to the question,
10 3
5832 = x(1− )
100
9 3
5832 = x( )
10
x = 8000
Thus, the purchased price = Rs. 8000

S13. Ans.(c)
Sol. Biology = 72%
Mathematics = 44%
Both = 72% + 44% – 100%
= 116% – 100%
= 16%
Here, 16 % → 40
40
1% →
16
40
100% → × 100
16
= 250

S14. Ans.(c)
Sol. Let the numbers of sum scored be x and n number of wickets.
x
= 12.4 ⇒ x = 12.4 n
n
According to the question,
12.4n+26
= 12.2
n+5
12.4n + 26 = 12.2n + 61
0.2n = 35
n = 175

S15. Ans.(b)
Sol. Quantity Price
I → 120 Rs. 360
II → 120 Rs. 240
240 Rs.600
Selling price of 25 → Rs. 60
Selling price of 240 → Rs. 576
Now, Loss = 600 – 576 = 24
Loss %= 4%

S16. Ans.(c)
Sol.

Here, 7r corresponds to 7 technician.
So, 15r → 15.
Total no. of workers = 22

S17. Ans.(b)
A + B
Sol. = 20 ⇒ A + B = 40
2
B + C
= 19 ⇒ B + C = 38
2
C + A
= 21 ⇒ C + A = 42
2
2(A + B + C) = 120
A + B + C = 60
A → 60 – 38
A = 22

S18. Ans.(c)
1
Sol. a + b + c = 1 and ab + bc + ca =
3
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
1
1 = a² + b² + c² + 2 ×
3
1
a² + b² + c² =
3
1
Now, a² + b² + c² = ab + bc + ca =
3
Thus, a = b = c
a : b : c = 1 : 1 : 1

S19. Ans.(b)
Sol.
Rs.1 : 50p : 25p
 8 : 5 : 3
value Rs.8 : Rs.2.50 : 0.75
Total value → 11.25
According to the question,
This,
11.25r → Rs. 112.50
1r → 10
No. of 50 paise coin = 50

S20. Ans.(b)
Sol. Let the no. of boys = 100
Girls = 120
B : G
Ratio → 5 : 6
According to the question,
11r → 66
1r → 6
Girls → 36, Boys → 30
30
Ratio → = 3 : 4
40

S21. Ans.(a)
Sol.
Milk : Water
VesselI  7 : 5
VesselII  17 : 7
Mixture  5 : 3
Let the capacity of container = 48l.

S22. Ans.(b)
Sol.
Shareof Milk : Shareof Water
4 9 1 11
3 2 3 2
5 20 5 20
:
5 5

33 : 17

S23. Ans.(a)
Sol. Total cost = 80 × 13.50 + 120 × 16
Cost of 200 kg = 3000
Cost of 1 kg → Rs. 15
Required profit = 20%
So, S.P = 15(1.2)
Then, selling price = Rs. 18/kg

S24. Ans.(a)
20 10 5
Sol. S.P = M.P (1− )(1− )(1− )
100 100 100
4 9 19
= 2000× × ×
5 10 20
= 1368

S25. Ans.(c)
Sol. Let the original price be Rs. 100
C.P → Rs. 80
S.P → 80 (1.4)
= 112
Profit % = 12%

S26. Ans.(d)
Sol.
C.P. of Rita = 16800 – 800
= 16000
80
M.R.P.× = 16000
100
M.R.P. = 20,000 Rs

S27. Ans.(d)
1
Sol. Rebate = 25% =
4
This rebate of Rs. 80 is on 1 shirt.
Rebate of Rs. 400 is on 5 shirts.
MP SP
4 : 3

4r → 320
1r → 80
3r → 240

S28. Ans.(c)
Sol. Let the CP be Rs. x.
According to the questions,
(0.75x) (1.32) = (1.15x) – 60
0.99x = 1.15x – 60
0.16x = 60
x = 375

S29. Ans.(a)
Sol.

Here, 165r → 330
1r → 2
Cost of A = Rs. 200
S30. Ans (c)
Sol.
Let price of each Article is Rs 1
1ST TRANSACTION
CP of 100 articles = 100
SP of 50 Articles= 60 [Profit % 20]
SP of 50 Articles=70 [Profit % 40]
Total SP = 130
2ND TRANSACTION
CP OF 100 ARTICLES = 100
SP OF 100 ARTICLES = 125[PROFIT % 25]

DIFFERENC IN PROFIT = 130-125=5
5 UNIT = RS100
1 UNIT = RS 20

S31. Ans.(b)
Sol.

95 (1.1)
MP → = 110
0.95
Thus, marked price = Rs. 110

S32. Ans.(a)
96 140
Sol. Total cost price of Tea = +
(0.8) (1.25)
= 120 + 112
Cost price of 2 kg = 232
CP of 1 kg = 116
S.P of 1 kg = 174
58
Gain % = ×100
116
= 50%

S33. Ans.(c)
Sol.
C.P S.P Profit Profitin(Rs.)
I 100 120 20%(onCP) 20
II 96 120 20%(onSP) 24

Now, difference of their profit = 4
Here, 4r → Rs. 85
85
120 → × 120
4
= 2550

S34. Ans.(d)
Sol.
Let the expenses be Rs. 120, 170 and 30.
Total expense = 320.
Now, increment in the respectively expenses

increament = 410 – 320 = 90
90 1
% increament = ×100 = 28 %
320 8

S35. Ans.(b)
Sol. 130% → Rs. 1326
100% → 1020
Thus, principal = Rs. 1020
Let the amount in t years = 1530
Now, Interest = 1530 – 1020
= 510
So,
5
510 = 1020 = × t
100
t = 10 years

S36. Ans.(b)
Sol.
Due debt×100 6450×100
Installment = = = 1500 Rs.
r×t×(t−1) 5×4×3
100t+ 400+
2 2

S37. Ans.(b)
Sol.
P×r×t
I =
100
Let the money lent at 5% be x & 4% be (60,000 – x)
According to the questions
x×5 4
2560 = +(60,000−x)×
100 100
x = 16,000
Money lent at 4% = 44,000

S38. Ans.(c)
Sol. Borrowed money at the rate = 4%
6
Lend money at the rate = % = (3%, half yearly)
2
t = 2 × 1 = 2 years
9
So, rate = 3 + 3 +
100
= 6.09%
Difference of rate = 2.09%
This 2.09% → 104.50
1% → 50
100% → 5,000
Thus, borrowed money = Rs. 5000

S39. Ans.(b)
2t
5
Sol. 9261 = 8000 (1+ )
100
9261 21 2t
= ( )
8000 20
21 3 21 2t
( ) = ( )
20 20
3
t =
2
= 1.5 years

S40. Ans.(a)
takes
Sol. Man → 30 hours
takes
Man + son → 20 hours
Let the work be L.C.M (30, 20) = 60 units

60
Time taken by son to finish alone = = 60 hours
1