Q4. What least number must be added to 1168, so the sum Q8.

Q8. A boat goes 11 km/h downstream and 5 km/h up-

is divisible by 26? stream. Find the speed in still water.
Solutions to Questions Solution: Find the remainder when 1168 is divided by 26: Solution: The speed in still water is the average of downstream and upstream speeds:

1168 mod 26 = 24 11 + 5
Speed in still water = = 8 km/h
2
The least number to be added is 26 − 24 = 2. Answer: 2
Answer: 8 km/h

Q1. The sum of ages of 5 children born at intervals of 3 Q5. A goods train runs at 72 km/h and crosses a 250 m
years is 50 years. What is the age of the youngest child? Q9. Simplify 63 + 83 − 73/62 + 83 − 72.
platform in 26 seconds. What is the length of the train?
Numerator:
Solution: Let the age of the youngest child be x. The ages of the 5 children will be: Solution: Convert speed to m/s: 63 + 83 − 73 = 216 + 512 − 343 = 385
x, x + 3, x + 6, x + 9, x + 12.
1000 Denominator:
x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50 72 km/h = 72 × = 20 m/s 62 + 83 − 72 = 36 + 512 − 49 = 499
3600
5x + 30 = 50 =⇒ 5x = 20 =⇒ x = 4 Total distance covered in 26 seconds: Result:
385
Answer: 4 years Distance = Speed × Time = 20 × 26 = 520 m 499
385
Answer: 499
Length of the train:
Q2. Three numbers are in the ratio 3:4:5 and their L.C.M. 520 − 250 = 270 m
is 2400. Find their H.C.F. Answer: 270 meters
Q10. Two trains meet; one reaches in 9 hours and the other
in 16 hours. Find the ratio of their speeds.
Solution: Let the numbers be 3x, 4x, 5x. The L.C.M. of these numbers is:
Q6. Three bells ring at intervals of 3, 5, and 7 seconds. Solution: Speed ratio is inverse of time taken after meeting:
LCM = HCF × Product of remaining factors = 2400
How many times do they ring together in 7 minutes? 16
The product of the ratios 3 × 4 × 5 = 60. Therefore: Ratio of speeds =
9
Solution: Find the LCM of 3, 5, and 7:
2400 Answer: 16:9
HCF = = 40
60 LCM = 105 seconds = 1.75 minutes
Answer: 40 In 7 minutes, they will ring together: Q11. A, B, and C finish a job in 4 days. Find time for C
7 alone.
Q3. The sum of two numbers is 25, and their difference is = 4 times (including the start at 0 seconds).
1.75
Solution: Work rates:
13. Find their product. Answer: 4 times 1
+
1 1
+ =
1
16 12 C 4
Solution: Let the two numbers be x and y. From the given conditions: 1
√ √ √ Solve for C
:  
Q7. Simplify ( 289/34) × (68/ 1156) × ( 961/62). 1 1 1 1 1 7 5
x + y = 25, x − y = 13 = − + = − =
C 4 16 12 4 48 48
Solution: √ √ √
Add and subtract to find: 289 = 17, 1156 = 34, 961 = 31 48
C= = 9.6 days
25 + 13 25 − 13 17 68 31 1 1 1 5
x= = 19, y= =6 × × = ×2× = Answer: 9.6 days
2 2 34 34 62 2 2 2
Product: Answer: 0.5
x × y = 19 × 6 = 114
Answer: 114

1 2 3

Q12. A and B complete work in 3 days with C; total pay- Q17. (Same as Q15) 4347
= (1.07)T − 1 =⇒ 1.07T = 1.1449
30000
ment = Rs. 3200. Find C’s share. Already solved above: Answer: 10 days From logs or approximation: T = 2 years. Answer: 2 years
Solution: Work rates:
1 1 4 3 7 Q18. Grocer’s sales for 5 months: Rs. 6435, Rs. 6927, Q23. Rs. 450 earns Rs. 81 interest at 4.5% p.a. Find the
A= , B = , A+B = + =
6 8 24 24 24
Work left for C: Rs. 6855, Rs. 7230, Rs. 6562. Sixth month to average Rs. time.
1 7 8 7 1
− = − = 6500? Solution: Simple Interest formula:
3 24 24 24 24
1
24 450 × 4.5 × T 81 × 100
C’s share = 1 × 3200 = 400. Answer: Rs. 400 Solution: Total sales for 6 months = 6 × 6500 = 39000. Sum of sales for 5 months =
3 81 = =⇒ T = = 4 years
6435 + 6927 + 6855 + 7230 + 6562 = 34009. Sixth month’s sales = 39000 − 34009 = 4991. 100 450 × 4.5
Answer: Rs. 4991 Answer: 4 years
Q13. A is thrice as good as a workman as B and therefore is
able to finish a job in 60 days less than B. Working together,
Q19. Average weight of A, B, C is 45 kg. A and B = 40 Q24. Town population increased from 1,75,000 to 2,62,500
they can do it in:
kg, B and C = 43 kg. Find B’s weight. in a decade. Average percent increase per year?
Solution: If A = x, B = 3x:
Solution: Let A = a, B = b, C = c. From the averages: - a + b + c = 45 × 3 = 135, - Solution: Formula for population growth:
x + 60 = 3x =⇒ x = 30 a + b = 40 × 2 = 80, - b + c = 43 × 2 = 86. From a + b + c = 135, subtract a + b:  r t
Time together: Pt = P0 1 +
1 1 4 1 c = 135 − 80 = 55 100
+ = =  r 10
30 90 90 22.5 From b + c = 86: 262500 = 175000 1 +
Answer: 22.5 days b = 86 − 55 = 31 100
262500  r 10
Answer: 31 kg = 1.5 = 1 +
175000 100
Q14. A, B, and C rates are given. Find time for B alone. Using logarithms or approximation:
Solution: Let A’s rate = 14 , B + C = 13 , and A + C = 12 : Q20. Average of nine consecutive odd numbers is 53. Find r ≈ 4 percent/year
1 1 1 1 the least number. Answer: 4%
= − =
B 3 4 12
B alone = 11 = 12 hours. Answer: 12 hours Solution: Let the numbers be x, x + 2, x + 4, . . . , x + 16. Average = middle number =
12
x + 8 = 53. Least number = x = 53 − 8 = 45. Answer: 45 Q25. Sugar price increased by 20%. What % decrease in
Q15. X and Y can do a piece of work in 20 days and 12 consumption to keep expenditure same?
Q21. Loan at 12% p.a. simple interest for 3 years; interest
days respectively. X started the work alone, and Y joined Solution: Let initial price = 100, consumption = 100 (expenditure = 10000). New
= Rs. 5400. Find principal. price = 120, let new consumption = x:
after 4 days. How long did the work last?
P ×R×T 10000
1 1
Solution: Simple Interest formula: SI = 100
. 120 × x = 100 × 100 =⇒ x = = 83.33
Solution: - X’s 1-day work = 20 , Y’s 1-day work = 12 . - Work done by X in 4 120
1 P × 12 × 3 5400 × 100
days = 4 × 20 = 15 . - Remaining work = 1 − 15 = 45 . - Combined rate of X and Y = Decrease = 100 − 83.33 = 16.67%. Answer: 16.67%
4 5400 = =⇒ P = = 15000
1 1 3 5 8 2 5 4 15 100 36
20
+ 12 = 60 + 60 = 60 = 15 . - Time to finish 45 work together = 2 = 5
× 2
= 6 days.
15
Total time = 4 + 6 = 10 days. Answer: 10 days Answer: Rs. 15,000
Q26. Average weight of 16 boys = 50.25 kg, 8 boys = 45.15
Q16. A pump fills a tank in 2 hours. Because of a leak, Q22. Compound interest on Rs. 30,000 at 7% p.a. = Rs. kg. Find overall average.
it took 2.5 hours. The leak drains the tank in how much 4347. Find the period. Solution: Total weight of 16 boys = 16 × 50.25 = 804, Total weight of 8 boys =
  8 × 45.15 = 361.2, Overall average:
time? R T
Solution: CI formula: CI = P 1 + 100
− P. 804 + 361.2 1165.2
1  T = = 48.55 kg
Solution: - Pump’s rate = 12 , effective rate (with leak) = 2.5 = 25 . - Leak’s rate = 7 16 + 8 24
1 5 4 1 4347 = 30000 1 + − 30000
2
− 25 = 10 − 10 = 10 . - Time to drain the tank = 11 = 10 hours. Answer: 10 hours 100 Answer: 48.55 kg
10

4 5 6
 4
Q27. Principal if CI - SI = 1134 at 15% p.a. for 3 years. Q3. Time trains meet: 5
Time = 2 = 6 days. Total = 4 + 6 = 10 days.
15
1 1
Distance = 110 km, speeds 20 km/h (Train A) and 25 km/h (Train B). (b) 10 women = 7 days → 1 woman = 70 . 10 children = 14 days → 1 child = 140
. 5
Solution: Difference formula:
Train A covers 20 km before Train B starts. Remaining distance = 110−20 = 90 km. women + 10 children:    
 2 1 1 1
R T Relative speed = 20 + 25 = 45 km/h. Time = 90 45
= 2 hours. Meeting time = 10 a.m. 5 + 10 =
CI - SI = P × R 70 140 10
100 1 + 100
Q4. Distance between Pune and Delhi: Time = 10 days.
Substitute values:  2
15 3 Let one train travel x km. Then the other travels x + 200. Speeds = 80 km/h, 95 km/h.
1134 = P × × Q8. Machines printing books:
100 1.15 Time is the same:
x x + 200
Solve for P : = Machine P = 8 hours, Q = 10 hours, R = 12 hours.
1134 × 1.15 80 95 From 9–11 a.m. (2 hours):
P = = 18000
0.0225 × 3 95x = 80(x + 200) =⇒ 15x = 16000 =⇒ x = 1066.67  
Answer: Rs. 18,000 1 1 1 37
Total distance = x + x + 200 = 2333.33 km. 2 + + =
8 10 12 60
23
Q28. A and B work 4 days; A and C complete in 4 more Q5. Four bells ringing together: Remaining work = 1 − 37
= 23
. Q + R: 11
. Time = 60
11 = 23
= 2.09 hours. Completion
60 60 60 60
11
days. C alone for 50% work? LCM of 4, 7, 12, 14 = 84 seconds. In 14 minutes (840 seconds): = 1 : 05 p.m..

Solution: Work done in 4 days: 840

= 10 times Q9. Population and averages:
84
4 4 1 2 5 (a) Male = x, Female = 8000 − x.
+ = + =
12 18 3 9 9 Q6. Work problems:
5 1.06x + 1.10(8000 − x) = 8600
Remaining work = 1 − 9
= 49 . A + C’s rate: (a) From the equations:
1 1 Solve: x = 4000, 8000 − x = 4000.
1 1 4
+ = =⇒ C = 36 days for full work. 6M + 8B = , 26M + 48B = (b) From averages:
12 C 9 10 2
36 1 1 A + B + C = 3 × 45 = 135, A + B = 2 × 40 = 80, B + C = 2 × 43 = 86
Time for 50% = 2
= 18 days. Answer: 18 days Solve: M = 100 , B = 200 .
Work by 15 men + 20 boys:
Solve: B = 31.
   
Detailed Solutions 1 1 1
15M + 20B = 15 + 20 =
100 200 4 Q10. Sums and percentages:
Q1. Ratio of present ages of father and son: 1
Time = 1 = 4 days. (a) Let x and y be two numbers:
4
From the equations: (b) From equations:
y − 10 = 3(x − 10) → y = 3x − 20 x + y = 5500, 1.1x + 1.2y = 6300
1 1
y + 10 = 2(x + 10) → y = 2x + 10 4M + 6W = , 3M + 7W = Solve: x = 2500, y = 3000. Difference = 3000 − 2500 = 500.
8 10 11628
Equating: (b) Total votes = 1136+7636+11628 = 20400. Winning candidate = 20400
×100 = 57%.
1 1
3x − 20 = 2x + 10 =⇒ x = 30, y = 70 Solve: M = 120 , W = 40 .
Work by 10 women:  
Ratio = 30 : 70 = 3 : 7. 1 1 Q11. Interest problems:
10W = 10 =
40 4 (a) From simple interest:
Q2. Sameer’s present age: 1
Time = 1 = 4 days. 5000 × r × 2 3000 × r × 4
4
+ = 2200
Let Sameer = 5x, Anand = 4x. From: 100 100
5x + 5 11 Q7. Work problems (continued): Solve: r = 10%.
= 4 14
4x + 5 9 (a) X = 20 days, Y = 12 days. X works 4 days → 20 = 15 . Remaining work = 1− 15 = 45 . (b) Numbers from 1 to 70 with 1 or 9 in units = 14. Percentage = 70
× 100 = 20%.
45x + 45 = 44x + 55 =⇒ x = 10 X + Y:
1 1 8 2
+ = =
Sameer’s age = 5x = 50 years. 20 12 60 15

7 8 9

Q12. Loans and averages:

(a) Interest = 5400, r = 12%, time = 3 years.
5400
Principal = = 15000
0.12 × 3
(b) Average of first 50 odd numbers:

Average = Middle term = 50

Q13. Compound interest:

(a) Difference = Re. 1, rate = 4%, time = 2 years:
1
Sum =   = 625
4 2
100

(b) CI on 8000:
 2
5
A = 8000 1 + = 8820
100

Q14. Work problems:

8 8
(a) X = 8 days → 40 . Remaining = 1 − 40
. Together = 10 days.
(b) B finishes work = 60.

10