Continued fraction
Start from the bottom: 6 + = .
Next: 5 + = 5 + = .
Finally: 5 + = 5 + .
= = — option (a).
54
A contractor employs 15 men to complete a work in 40 days, working 8 hours per day. After ten days he adds five more men but also reduces the working hours to 6 per day from the 11th day onwards. If all men work with the same efficiency throughout, how many more days will be required to finish the remaining work?
A.25
B.30
C.35
D.28
Solution
Men, days and hours
Total work = 15 × 40 × 8 = 4800 man-hours.
In the first 10 days: 15 × 10 × 8 = 1200 man-hours, leaving 3600.
From day 11 the daily output is 20 × 6 = 120 man-hours.
= 30 days — option (b).
55
Deepak covers half of his journey at 27 km/h and the remaining half at 9 km/h. His average speed is:
A.12.5 km/h
B.11.5 km/h
C.14.5 km/h
D.13.5 km/h
Solution
Average speed
For two equal halves, average speed = .
= .
= .
= 13.5 km/h — option (d).
56
The speed of a car is 126 km/h. Find the distance covered by the car in 60 seconds.
A.1900 m
B.2054 m
C.2100 m
D.2300 m
Solution
Distance in seconds
126 km/h = 126 × = 35 m/s.
Distance = speed × time.
= 35 × 60.
= 2100 m — option (c).
57
Partners A and B entered into a business investing ₹50,000 and ₹80,000 respectively. After 5 months C joins with ₹1,20,000. At the year-end the profit was ₹1,00,000. How much more did C earn than A?
A.₹6,000
B.₹11,000
C.₹10,000
D.₹14,000
Solution
Partnership
A: 50,000 × 12 = 6,00,000; B: 80,000 × 12 = 9,60,000; C: 1,20,000 × 7 = 8,40,000.
Ratio = 600 : 960 : 840 = 15 : 24 : 21, a total of 60 parts.
A gets × 1,00,000 = ₹25,000 and C gets × 1,00,000 = ₹35,000.
Difference = ₹10,000 — option (c).
58
A triangular field with base 70 m and height 96 m shares a side with a rectangle of length 65 m and breadth 35 m. Find the combined area and the percentage of the triangle's area in the total.
A.4,450 m² and 34.5%
B.3,450 m² and 25%
C.5,635 m² and 59.62%
D.5,500 m² and 59%
Solution
Combined area
Triangle = × 70 × 96 = 3360 m².
Rectangle = 65 × 35 = 2275 m².
Total = 3360 + 2275 = 5635 m².
× 100 ≈ 59.62% — option (c).
59
The area of an equilateral triangle increases by 44%. By what percentage did the side length increase?
A.30%
B.20%
C.34%
D.16%
Solution
Area and side
Area varies as the SQUARE of the side.
New area = 1.44 times the old, so the side ratio is = 1.2.
That is 1.2 times the old side.
An increase of 20% — option (b).
60
The weights of three friends are in the ratio 8 : 7 : 3. If each gains 9 kg, the new ratio becomes 17 : 16 : 12. What is the original weight of the heaviest friend?
A.4 kg
B.8 kg
C.10 kg
D.7 kg
Solution
Ratio after addition
Let the weights be 8x, 7x and 3x.
= ⇒ 128x + 144 = 119x + 153 ⇒ 9x = 9, so x = 1.
Check the third: = ✓.
Heaviest originally = 8 × 1 = 8 kg — option (b).
61
A triangle has vertices at A(8, 10), B(4, 4) and C(16, 6). What is its area?
The average of a certain number of quantities is 48. If 6 is added to each quantity, what will be the new average?
A.39
B.54
C.41
D.52
Solution
Average shift
Adding the same number to every quantity shifts the average by exactly that number.
So the new average = 48 + 6.
The count of quantities makes no difference.
= 54 — option (b).
63
A wholesaler sells a packet of tea to a retailer at a profit of 20%. The retailer then marks up the price by 35% and offers a discount of 16.67% to the customer. If the customer pays ₹864, find the cost price of the packet for the wholesaler.
A.₹600
B.₹400
C.₹640
D.₹750
Solution
Chain of profit and discount
16.67% = , so the discount leaves of the marked price.
Customer pays = CP × × × = CP × 1.35.
1.35 × CP = 864.
CP = = ₹640 — option (c).
64
A group of 7 students went on a trip and spent ₹2,500 in total. If one of them spent ₹700 while the rest spent equally, what was the average spending of the other 6 students?
A.₹300
B.₹360
C.₹500
D.₹240
Solution
Average spending
Total spent by the other six = 2,500 − 700 = ₹1,800.
They spent equally among themselves.
Average = .
= ₹300 — option (a).
65
In a quadrilateral ABCD, angles A, B and C are in the ratio 5 : 4 : 5. If angle D = 80°, what is angle B?
A.80°
B.72°
C.45°
D.120°
Solution
Angles of a quadrilateral
The four angles of a quadrilateral add up to 360°.
5x + 4x + 5x + 80° = 360° ⇒ 14x = 280°.
x = 20°.
Angle B = 4x = 80° — option (a).
66
A distributor marks a batch of 280 identical mobile chargers at 60% above the cost price and offers a 20% discount to retailers. A retailer purchases all the chargers and spends ₹15,360. What is the cost price per charger for the wholesaler?
A.₹42.85
B.₹57.75
C.₹56.65
D.₹55.55
Solution
Markup and discount
Take CP = 100 units ⇒ MP = 160; after a 20% discount SP = 160 × 0.8 = 128 units.
128 units = ₹15,360, so 100 units = = ₹12,000 — the total cost of all 280.
Cost per charger = .
≈ ₹42.85 — option (a).
67
Three numbers are in the ratio 4 : 5 : 7. If their average is 480, what is the difference between the largest and the smallest number?
A.270
B.300
C.240
D.250
Solution
Ratio and average
Sum of the three = 3 × 480 = 1440.
4x + 5x + 7x = 16x = 1440, so x = 90.
Largest = 7 × 90 = 630 and smallest = 4 × 90 = 360.
Difference = 270 — option (a).
68
A 136-litre container of pure milk undergoes a process in which 17 litres is removed and replaced with water. The process is carried out 3 times in all. Approximately how much of the original milk remains?
A.17 litres
B.17 litres
C.81 litres
D.91 litres
Solution
Repeated replacement
Each round leaves (1 − ) = of the milk behind.
After three rounds the milk left = 136 × ()³.
= 136 × = 17 × .
= = 91 litres — option (d).
69
If secx − tanx = , then what is the value of secx + tanx?
A.9
B.3
C.4
D.1
Solution
sec and tan identity
The identity sec²x − tan²x = 1 gives (secx − tanx)(secx + tanx) = 1.
So secx + tanx is the reciprocal of secx − tanx.
= 1 ÷ .
= 3 — option (b).
70
Two circles of radii 8 cm and 5 cm are such that the distance between their centres is 13 cm. How many common tangents can be drawn?
A.0
B.1
C.2
D.3
Solution
Circles touching externally
Here d = 13 and r₁ + r₂ = 8 + 5 = 13, so d = r₁ + r₂.
The circles therefore touch each other externally at exactly one point.
In this case there are two direct common tangents plus one at the point of contact.
Total 3 — option (d).
In a circle, the chords EF and GH are equal. The angle subtended by EF at the circumference is 95°. What is the angle subtended by GH at the circumference?
A.95°
B.47.5°
C.120°
D.85°
Solution
Equal chords
Equal chords of a circle subtend equal angles at the centre.
Half of those equal central angles are the angles at the circumference.
So equal chords subtend equal angles at the circumference as well.
= 95° — option (a).
74
The number of students in two sections A and B having different heights is shown in the table below.
Height 1.35 m — A: 7, B: 6
Height 1.40 m — A: 9, B: 8
Height 1.45 m — A: 10, B: 12
Height 1.50 m — A: 13, B: 11
Height 1.55 m — A: 3, B: 2
Height 1.60 m — A: 12, B: 6
Height 1.65 m — A: 1, B: 4
What is the ratio of the total number of students in Section A whose height is less than 1.50 m to the total number of students in Section B whose height is more than 1.50 m?
A.15 : 16
B.13 : 6
C.13 : 5
D.17 : 13
Solution
Reading a table
Section A below 1.50 m: 7 + 9 + 10 = 26.
Section B above 1.50 m: 2 + 6 + 4 = 12.
Ratio = 26 : 12.
= 13 : 6 — option (b).
75
Which of these lines is parallel to 3x − 2y = 7?
A.3x + 2y = 7
B.4x + y = 15
C.3x − 2y = 11
D.x + 5y = 10
Solution
Parallel lines
Two lines are parallel when their slopes are equal.
That happens when the coefficients of x and y are in the same proportion.
3x − 2y = 11 has exactly the same coefficients as the given line, only the constant differs.
Hence option (c).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Write the condition as = = = k.
Then A = 2k, B = 4k and C = 6k.
A : B : C = 2 : 4 : 6.
= 1 : 2 : 3 — option (a).
Innermost bracket: − = = ; then × = .
Curly bracket: − = = .
Square bracket: ÷ = × 12 = 10.
− 10 = = − — option (a).
Start from the bottom: 6 + = .
Next: 5 + = 5 + = .
Finally: 5 + = 5 + .
= = — option (a).
Total work = 15 × 40 × 8 = 4800 man-hours.
In the first 10 days: 15 × 10 × 8 = 1200 man-hours, leaving 3600.
From day 11 the daily output is 20 × 6 = 120 man-hours.
= 30 days — option (b).
For two equal halves, average speed = .
= .
= .
= 13.5 km/h — option (d).
126 km/h = 126 × = 35 m/s.
Distance = speed × time.
= 35 × 60.
= 2100 m — option (c).
A: 50,000 × 12 = 6,00,000; B: 80,000 × 12 = 9,60,000; C: 1,20,000 × 7 = 8,40,000.
Ratio = 600 : 960 : 840 = 15 : 24 : 21, a total of 60 parts.
A gets × 1,00,000 = ₹25,000 and C gets × 1,00,000 = ₹35,000.
Difference = ₹10,000 — option (c).
Triangle = × 70 × 96 = 3360 m².
Rectangle = 65 × 35 = 2275 m².
Total = 3360 + 2275 = 5635 m².
× 100 ≈ 59.62% — option (c).
Area varies as the SQUARE of the side.
New area = 1.44 times the old, so the side ratio is = 1.2.
That is 1.2 times the old side.
An increase of 20% — option (b).
Let the weights be 8x, 7x and 3x.
= ⇒ 128x + 144 = 119x + 153 ⇒ 9x = 9, so x = 1.
Check the third: = ✓.
Heaviest originally = 8 × 1 = 8 kg — option (b).
Area = |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
= |8(4 − 6) + 4(6 − 10) + 16(10 − 4)|.
= |−16 − 16 + 96| = × 64.
= 32 square units — option (a).
16.67% = , so the discount leaves of the marked price.
Customer pays = CP × × × = CP × 1.35.
1.35 × CP = 864.
CP = = ₹640 — option (c).
Total spent by the other six = 2,500 − 700 = ₹1,800.
They spent equally among themselves.
Average = .
= ₹300 — option (a).
The four angles of a quadrilateral add up to 360°.
5x + 4x + 5x + 80° = 360° ⇒ 14x = 280°.
x = 20°.
Angle B = 4x = 80° — option (a).
Take CP = 100 units ⇒ MP = 160; after a 20% discount SP = 160 × 0.8 = 128 units.
128 units = ₹15,360, so 100 units = = ₹12,000 — the total cost of all 280.
Cost per charger = .
≈ ₹42.85 — option (a).
Sum of the three = 3 × 480 = 1440.
4x + 5x + 7x = 16x = 1440, so x = 90.
Largest = 7 × 90 = 630 and smallest = 4 × 90 = 360.
Difference = 270 — option (a).
Each round leaves (1 − ) = of the milk behind.
After three rounds the milk left = 136 × ()³.
= 136 × = 17 × .
= = 91 litres — option (d).
The identity sec²x − tan²x = 1 gives (secx − tanx)(secx + tanx) = 1.
So secx + tanx is the reciprocal of secx − tanx.
= 1 ÷ .
= 3 — option (b).
Here d = 13 and r₁ + r₂ = 8 + 5 = 13, so d = r₁ + r₂.
The circles therefore touch each other externally at exactly one point.
In this case there are two direct common tangents plus one at the point of contact.
Total 3 — option (d).
Degrees = radians × .
= × .
= 3 × 90.
= 270° — option (c).
SI = .
= .
= .
= ₹1,800 — option (c).
Section A below 1.50 m: 7 + 9 + 10 = 26.
Section B above 1.50 m: 2 + 6 + 4 = 12.
Ratio = 26 : 12.
= 13 : 6 — option (b).