1. Ratio of the time taken by A alone to complete a certain piece of work while that by B and C to complete the same work together is 4 : 3. If the time taken by C to complete the work alone is 8 days and the ratio of the efficiencies of B and C is 1 : 3, then find the time in which A and B can complete the work while working together.
A. 4.5 days
B. 6 days
C. 7.5 days
D. 8 days
2. The length of each side of a rhombus is 29 cm while that of one of the diagonals is 40 cm. What will be the area of the rhombus?
A. 840 [latex]{cm}^{2}[/latex]
B. 880 [latex]{cm}^{2}[/latex]
C. 820 [latex]{cm}^{2}[/latex]
D. 800 [latex]{cm}^{2}[/latex]
3. Three successive discounts of 15%, 12.5%, and 20% are equivalent to a single discount of
A. 38.5 %
B. 41.5 %
C. 39.8 %
D. 40.5 %
4. A and B together have Rs.2,448. If 25 % of A's amount is equal to 35 % of B's amount, then find the amount with A.
A. Rs. 1,020
B. Rs. 1,428
C. Rs. 1,462
D. Rs. 1,468
5. If the cost price of 50 articles is equal to the selling price of 40 articles, then find the profit or loss percent.
A. 20 %
B. 16.67 %
C. 25 %
D. 33.33 %
6. A’s income is 25% more than B’s income while C’s income is 20% more than B's income. By what percent is A’s income more or less than C's income?
A. 5.26%
B. 4.16%
C. 4.28%
D. 4.46%
7. In what time will a 143 m long train running at the rate of 45 km/hr cross a man coming towards it at the rate of 4.5 km/hr?
A. 11.4 sec
B. 10.4 sec
C. 11.64 sec
D. 10.64 sec
8. B : A + [latex]\sqrt{({A}^{2} + {B}^{2})}[/latex] = 2 : 3 then find the value of B : A
A. 8 : 15
B. 24 : 7
C. 12 : 5
D. 6 : 8
9. If [latex]({cos}^{2} A - sin A)(sin A + {cos}^{2} A)[/latex] = -1, then find the value of 2 + [latex]{cos}^{2} A[/latex].
A. 0
B. [latex]{sec}^{2} A[/latex]
C. 1
D. 2
10. If four interior angles of a pentagon are 140°, 90°, 70° and 80°, then find the value of the fifth interior angle.
A. 20°
B. 160°
C. 140°
D. 40°
11. Find the value of [latex]secA(secA + tanA + \frac{cos A}{1 + sinA}) - 2{tan}^{2} A[/latex]
12. Find the value of [latex]\frac{2 * {8}^{n + 1} + 4 * {2}^{3n - 1} }{4 * {2}^{2n + 1} - 2 * {4}^{3n + 1} } [/latex]for n = 2.
A. [latex]\frac{3}{4}[/latex]
B. [latex]- \frac{3}{4}[/latex]
C. [latex] \frac{3}{5}[/latex]
D. [latex]- \frac{3}{5}[/latex]
13. If the average age of sixteen students is 18.5 years and the average age of these students along with their teacher is 22 years, then find the age of the teacher?
A. 58 years
B. 78 years
C. 68 years
D. 76 years
14. If b + [latex]\frac{1}{b} = -1,[/latex] then find the value of [latex]{b}^{9} + \frac{1}{{b}^{15}}[/latex]
15. Medians BE and CF of △ABC intersects at point O. P and Q are the midpoints of BO and CO respectively. If PQ = 3.5 cm, then find the length of BC.
A. 3.5 cm
B. 7 cm
C. 10.5 cm
D. 4.5 cm
16. In triangle PQR, S and T are the points on PQ and PR respectively, such that ST is parallel to QR and PS : SQ = 3 : 2. If RT = 5 cm, then find the ratio of the areas of triangle PST and quadrilateral STRQ.
A. 9 : 4
B. 9 : 25
C. 16 : 25
D. 9 : 16
17. If Rs. 14,600 amounts to Rs. 16,404.56 invested in compound interest (compounded annually) for two years, then the rate of interest is
A. 5 %
B. 6 %
C. 7 %
D. 8 %
18. From the top of a tower 40[latex]\sqrt{3} [/latex] meters high the angle of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. If the pole and tower stand on the same plane, then find the distance between the top of the tower and that of the pole
A. [latex]20 \sqrt{3} [/latex]
B. [latex]20 \sqrt{2} [/latex]
C. [latex]40 \sqrt{2} [/latex]
D. 8.6 m
19. A person borrows Rs. 33,000 at 20% compound interest. How much he has to pay equally at the end of each year, to settle his loan in two years?
A. Rs. 22, 450
B. Rs. 21, 600
C. Rs. 17, 000
D. Rs. 19, 600
20. A faulty clock shows correct time at 6 a.m. It lags 1sec at the end of every 10 sec. What will be the time shown by the clock at 9 p.m.?
A. 7:00 p.m.
B. 7:30 p.m.
C. 8:00 p.m.
D. 8:30 p.m.
21. Given that [latex]{x}^{2} - {y}^{2}[/latex] = 60. If x = 6 + y, then the average of x and y is
22. What is the value of x, if x = [latex]\frac{1}{3 + \frac{3}{5 + \frac{4}{5 + \frac{7}{2}}}} = ?[/latex]
A. [latex]\frac{31}{110}[/latex]
B. [latex]\frac{31}{101}[/latex]
C. [latex]\frac{41}{101}[/latex]
D. [latex]\frac{3}{11}[/latex]
23. If [latex]\frac{a}{b} = \frac{1}{4} and \frac{b}{c} = \frac{2}{3} , then \frac{12 {a}^{2} - 8 {c}^{2}}{33 {a}^{2} + {c}^{2}}[/latex] is equal to
A. -4
B. -1
C. 0
D. [latex]\frac{2}{17}[/latex]
24. O is the in center of the ΔABC. From point O a perpendicular is drop on side BC such that
perpendicular meets BC at point P. If ∠BOP 39, then ∠ABC is
A. 39°
B. 78°
C. 51°
D. 102°
25. The difference of a number consisting of three different digits from the number formed by reversing the digits is always divisible by
A. 9
B. 10
C. 11
D. Both (a) and (c)
26. Pipe A can empty a tank in 12 minutes, pipe B can empty the tank in 18 minutes while the pipe C can empty the tank in 36 minutes. In how much time the tank will be empty if all three pipes are opened together?
A. 22 minutes
B. 10 minutes
C. 6 minutes
D. 5 minutes
27. A circle is inscribed in an equilateral triangle of side 6 cm. A square is inscribed in this circle, then the area of the square (in [latex]{cm}^{2}[/latex]) is
28. A conical vessel of base radius 9 cm and height 8 cm is filled with milk. If the milk leaks through a hole at the bottom of the conical vessel into a cylindrical jar of radius 3 cm, then find the level of the milk in the jar after the milk has leaked completely.
A. 12 cm
B. 15 cm
C. 24 cm
D. 27 cm
29. A shopkeeper sells an article at [latex]12 \frac{1}{2}[/latex] % loss. If he sells it for Rs. 51.80 more, then he gains 6 %. What is the cost price of the article?
A. Rs. 210
B. Rs. 240
C. Rs. 280
D. Rs. 300
30. Ramesh traveled from city A to city B, in which he traveled 42 km by motorcycle and [latex]{\frac{1}{4}}^{th}[/latex] of the total distance by car. If the rest 40% of the distance is traveled in a bus, then what is the total distance?
A. 100 Km
B. 120 Km
C. 140 km
D. 160 km
Answers and Explanations
1. Answer - Option B
Explanation -
Let the total work be 8 × 3 = 24 units.
[latex]\Rightarrow[/latex] Work done by C in 1 day = 3 units and that by B = 1 unit.
Thus, time taken by B and C together = 6 days while that by A alone = 8 days.
[latex]\Rightarrow[/latex] Work done by A in 1 day = 3 units
Hence, required time = [latex]\frac{24}{4}[/latex] = 6 days.
2. Answer - Option A
Explanation -
[latex]{AB}^{2} = {AO}^{2} + {OB}^{2}[/latex]
[latex]\Rightarrow {29}^{2} = {20}^{2} + {OB}^{2}[/latex]
[latex]\Rightarrow OB = 21 cm[/latex]
Hence, required area
[latex]= 4 * \frac{1}{2} * 21 * 20 = 840{cm}^{2}[/latex]
3. Answer - Option D
Explanation -
[latex]= x * \frac{17}{20} * \frac{7}{8} * \frac{4}{5} = * \frac{119}{200} =[/latex] 59.5% of x
Hence, required discount = 40.5%.
4. Answer - Option B
Explanation -
25% of A = 35% of B
[latex]\Rightarrow[/latex] A : B = 7 : 5
Hence, required amount = [latex]\frac{7}{12}[/latex] × 2448
= Rs. 1,428
5. Answer - Option C
Explanation -
Let the selling price of 1 article be Re.1
[latex]\Rightarrow[/latex] Cost price of 50 articles = Rs. 40 and their
selling price = Rs. 50
Hence, there would be profit equivalent to
[latex]\frac {50 – 40}{40}[/latex] × 100 = 25%
6. Answer - Option B
Explanation -
Let the income of B = Rs. 100
[latex]\Rightarrow[/latex] A's income = Rs. 125 and C's income = Rs. 120
Hence, required percentage
= [latex]\frac {125 – 120}{120}[/latex] × 100 = 4.16%
7. Answer - Option B
Explanation -
Relative speed = 45 + 4.5 = 49.5 km/hr
= 49.5 × [latex]\frac{5}{18}[/latex] = 13.75 m/s
Hence, required time = [latex]\frac{143}{13.75}[/latex] = 10.4 seconds.
8. Answer - Option C
Explanation -
Going through the options we get the required answer as 12 : 5.
9. Answer - Option
Explanation -
[latex]({cos}^{2} A - sin A)(sin A + {cos}^{2} A)[/latex] = -1
[latex]\Rightarrow {cos}^{4} A - {sin}^{2} A[/latex] = -1
[latex]\Rightarrow {cos}^{4} A + {cos}^{2} A[/latex] = 0
[latex]\Rightarrow {cos}^{2} A({cos}^{2} A + 1)[/latex] = 0
Since [latex]{cos}^{2} A [/latex] can never be negative in the given range thus A = 90° and [latex]({cos}^{2} A + 2)[/latex] = 2
10. Answer - Option B
Explanation -
Let the fifth angle be x°.
Sum of all the exterior angles of any polygon is always 360°.
[latex]\Rightarrow[/latex] (180 – 140)° + (180 – 90)° + (180 – 70)° + (180 – 80)° + (180 – x)° = 360°
Hence, x = 160°.
11. Answer - Option C
Explanation -
[latex]secA(secA + tanA + \frac{cos A}{1 + sinA}) - 2{tan}^{2} A[/latex], we get the required value as 2.
12. Answer - Option D
Explanation -
[latex]\frac{2 * {8}^{n + 1} + 4 * {2}^{3n - 1} }{4 * {2}^{2n + 1} - 2 * {4}^{3n + 1} } [/latex]
= [latex]\frac{{2}^{3n + 4} + 2 * {2}^{3n + 1} }{{2}^{2n + 3} - {2}^{4n + 3} } [/latex]
= [latex]\frac{{2}^{10} + 2 * {2}^{7} }{{2}^{7} - {2}^{11} } = - \frac {3}{5}[/latex]
13. Answer - Option B
Explanation -
Age of the teacher = 17 × 22 – 16 × 18.5
= 78 years.
14. Answer - Option A
Explanation -
b + [latex]\frac{1}{b} = -1[/latex]
[latex]\Rightarrow {b}^{2} + b + 1[/latex]
[latex]\Rightarrow (b - 1)({b}^{2} + b + 1) = 0[/latex]
[latex]\Rightarrow {b}^{3} = 1[/latex]
[latex]{b}^{9} + \frac{1}{{b}^{15}} = 2[/latex]
15. Answer - Option
Explanation -
Using mid-point theorem:
BC = 7 cm.
16. Answer - Option
Explanation -
Required ratio
[latex]\frac{Area of △ PST}{Area of △ STRQ} = \frac{{(3x)}^{2}}{{(5x)}^{2} - {(3x)}^{2}} = \frac{9}{16}[/latex]
17. Answer - Option B
Explanation -
14600 * [latex]{(1 + \frac{r}{100})}^{2}[/latex] = 16,404.56
[latex]\Rightarrow [/latex]r = 6 %
18. Answer - Option C
Explanation -
In △ABE ,
AB tan 30º = [latex]\frac{AB}{40 \sqrt{3}}\Rightarrow AB = 40 m[/latex]
In △EDC,
sin 45º [latex]\frac{40}{EC}\Rightarrow EC = 40 \sqrt{2} m[/latex]
19. Answer - Option B
Explanation -
[latex](33000(1 +\frac{20}{100 ) - x (1 + \frac{20}{100})} = x[/latex]
[latex]\Rightarrow \frac{x * 5}{6} + x = 33000 * \frac{6}{5}[/latex]
[latex]\Rightarrow x * \frac{11}{6} = 39600[/latex]
[latex]\Rightarrow x = Rs. 21,600[/latex]
20. Answer - Option B
Explanation -
Difference between the two time limits
= 15 hrs. = 15 × 60 × 60 = 54000 sec
[latex]\Rightarrow[/latex] Lag at the end of 54000 sec
= 5400 sec = 1 hr 30 min
Hence, the time shown = 7:30 p.m.
21. Answer - Option C
Explanation -
Given x – y = 6 and [latex]{x}^{2} - {y}^{2}[/latex] = 60
therefore, x + y= 10
Hence, the average of x and y = [latex]\frac{x + y}{2} = \frac{10}{2} = 5[/latex]
22. Answer - Option A
Explanation -
x = [latex]\frac{1}{3 + \frac{3}{5 + \frac{4}{5 + \frac{7}{2}}}}[/latex]
x = [latex]\frac{1}{3 + \frac{3}{5 + \frac{4}{\frac{17}{2}}}}[/latex]
x = [latex]\frac {1}{3 + \frac{3}{\frac{93}{17}}}[/latex]
x = [latex]\frac{1}{3 + \frac{51}{93}}[/latex]
x = [latex]\frac{93 }{31} = \frac{31 }{110}[/latex]
23. Answer - Option A
Explanation -
[latex]\frac{a}{b} = \frac{1}{4} and \frac{b}{c} = \frac{2}{3}[/latex]
[latex]\Rightarrow a : b : c = 1 : 4 : 6[/latex]
[latex]\frac{12 {a}^{2} - 8 {c}^{2}}{33 {a}^{2} + {c}^{2}} = \frac{12 {1}^{2} - 8 {6}^{2}}{33 {1}^{2} + {6}^{2}}[/latex]
[latex]\frac {12 (1 - 24)}{3 (11 + 12)} = -4[/latex]
24. Answer - Option D
Explanation -
∠ABO = ∠OBP
= [latex]\frac{∠ABC}{2}[/latex]
and ∠OBP = 90º - ∠BOP
= 90º - 39º - 51º
Hence, ∠ABC = 2 * 51º = 102º.
25. Answer - Option D
Explanation -
Let the 3 digit number be 100x + 10y + z. Then,
(100z + 10y + x) – (100x + 10y + z)
= 99(z – x) = 9 × 11 × a number.
26. Answer - Option C
Explanation -
Let the capacity of the tank be (LCM of 12, 18 and 36) 36 units. Then,
Pipe A empties [latex]\frac{36}{12}[/latex] = 3 units,
pipe B empties [latex]\frac{36}{18}[/latex] = 2 units
while pipe C empties [latex]\frac{36}{36}[/latex] = 1 unit from the tank in 1 minute.
Hence, required time = [latex]\frac{36}{3 + 2 + 1}[/latex] = 6 minutes.
27. Answer - Option B
Explanation -
Let ‘r’ be the radius of the circle and ‘a’ be the length of the side of the square. Then, ‘r’ is the
in radius of the equilateral triangle.
[latex]\Rightarrow r = \frac{6}{2 \sqrt{3}} = \sqrt{3} cm[/latex]
[latex]\Rightarrow [/latex] Diameter= 2r = 2 [latex] \sqrt{3} cm[/latex]
i.e, Diameter of the circle = Diagonal of the square
[latex]\Rightarrow [/latex] Diagonal of the square = 2 [latex] \sqrt{3} cm[/latex]
[latex]\Rightarrow [/latex] Side of the square = [latex] \frac{2 \sqrt{3}}{\sqrt{2}} = \sqrt{6} cm[/latex]
Hence, area of the square = [latex]{\sqrt{6}}^{2} = 6 {cm}^{2}[/latex]
28. Answer - Option C
Explanation -
Let the level of the milk in the jar be ‘h’. Then,
volume of the conical vessel = volume of the cylinder jar
[latex]\Rightarrow \frac{1}{3} \pi {(9)}^{2} * 8 = \pi {(3)}^{2} * h[/latex]
[latex]\Rightarrow [/latex] h = 24 cm
29. Answer - Option C
Explanation -
Let the cost price be Rs. x. Then,
[latex]\frac{(12.5 + 6)x}{100} = 51.80[/latex]
[latex]\Rightarrow [/latex] x = 280
30. Answer - Option B
Explanation - Option
Let the distance between city A and city B be x km. Then,
[latex]42 + \frac{x}{4} + \frac{2x}{5} = x[/latex]
[latex]\Rightarrow 42 + \frac{(5 + 8)x}{20} = x[/latex]
[latex]\Rightarrow[/latex] x = 120