Monday, 3 September 2018



कार्यालय सहायक II 2018

 OFFICE ASSISTANT CLERK GRADE II MATH SOLUTIONS

101. In the given figure, if BD||RS||PQ, CP=PD=11cm, AR=RD=3 cm, BD=x, RS=y, PQ=z, then the ratio of values of y and z is?
दिए गए आंकड़े में, यदि बीडी || आरएस || पीक्यू, सीपी = पीडी = 11 सेमी, एआर = आरडी = 3 सेमी, बीडी = एक्स, आरएस = वाई, पीक्यू = जेड, फिर वाई और जेड के मूल्यों का अनुपात है?








(A) 1:2
(B) 6:11
(C) 3:11
(D) 1:1
D
∠A is common AR/AD = 3/3
∠ARS is congruent to ∠ADB
RS/DB = AR/AD and RS/x = 3/6 = ½ = y/x
Similarly Q/x = 11/22 =1/2 = z/x
Therefore the ratio of values of y and z are 1:1


104. The single discount which is equivalent to discount of 10%, 20% and 25% is
एकल छूट 10%, 20% और 25% छूट ...... के बराबर है
(A) 45%
(B) 55%
(C) 46%
(D) 54%
Marked price = 100
after 10% discount  = (100-10)=90
after 20% discount on 90 = (90-18)=72
after 25 % discount on 72 = (72-18) = 54
Total discount = 10+18+18 = 46%
 (0.75/1)×(0.80/1)×(.90/1) = 0.54/1
Marked value  Rs 1 and sale value = 0.54 therefore discount =1-0.54 =0.46×100=46%
105. A sum of money doubles itself at compound interest in 5 years. It will be eight times in : 
5 साल में चक्रवृधि ब्याज पर धनराशि खुद ही दोगुना हो जाती है, तो आठ बार होगी ?

(1) 20 years 

(2) 10 years 

(3) 12 years 
(4) 15 years 

4
P×2 in five years
P×4 in ten years
P×8 in fifteen years
 
106. Volume of a cone is 16632 cm3 and height is 9cm What is its base radius ? (∏= 22/7)
एक शंकु का आयतन 16632 सेमी3 है और ऊंचाई 9 सेमी है इसकी बेस त्रिज्या क्या है? (Π = 22/7)
(A) 48 cm
(B) 36 cm
(C) 42 cm
(D) 49 cm
C
शंकु का आयतन 1/3∏r2h
16632 = 1/3 × 22/7×r²×9
r²×3×22/7 = 16632
r²×66/7 = 16632
r² = 16632×7/66 =1764
r=√1764 = 42

107.  Volume of a cube is 128√2 cm3, What is the area of one face of the cube
घन का वॉल्यूम 128√2 सेमी3 है, घन के एक चेहरे का क्षेत्र क्या है
(A) 48 cm2
(B) 32 cm2
(C) 16 cm2
(D) 40 cm2
B
Volume of cube = a3
a3 =128√2
a =3√(128√2)
a6  =1282  × 2 = 16384×2, = 32768
a2  = 3√32768 = 32
area of one face = 32 cm2

108. PQ is a straight line of 13 units length. If P has the co-ordinates (2,5) and Q has the co-ordinates (x, -7) then the value of x is
पीक्यू 13  इकाइयों की लंबाई की सीधी रेखा है। यदि पी में समन्वय (2,5) है और क्यू में समन्वय (x, -7) है तो x का मान है
(A) 7
(B) -7
(C) 3
(D) 13
C

 (
(x+2)2 + 122 = 132 
(x+2)2 =169-144 = 25
 (x)2 =25-(2)2
(x)2 =25-16 = 9

x=3


109. Which of the following points lie on x- axis ?
निम्नलिखित में से कौन सा बिंदु एक्स-अक्ष पर है?
(a) (-2, 0),  (b) (-2, -2) , (c) (0, -2)
(A) (-2, 0)
(B) (-2, -2)
(C) (0, -2)
(D) (2,2)
A
(-2, 0) = (-2) on x and  0 on x
(-2, -2)=  (-2) on x and  -2 on y
(0, -2) = (0) on x and  -2 on y
(-2, 0) एक्स-अक्ष पर है?

110. If the co-ordinates of three vertices of a square are (0,0), (0,-4), and (4,0), then the co-ordinates of its fourth vertex is –
यदि वर्ग के तीन शीर्षकों के समन्वय (0,0), (0, -4), और (4,0) हैं, तो इसके चौथे शीर्षक का समन्वय हैं –
(A) (-4, -4)
(B) (4, 4)
(C) (-4, 4)
(D) (4, -4)

D






यदि वर्ग के तीन शीर्षकों के समन्वय (0,0), (0, -4), और (4,0) हैं, तो इसके चौथे शीर्षक का समन्वय होगा (4, -4)

111.  The length of the shadow of a vertical tower on level ground increase by 10m, when the angle of elevation of the sun change from 45º to 30º. The height of the tower is  -
जमीन स्तर पर एक ऊर्ध्वाधर टावर की छाया की लंबाई 10 मीटर तक बढ़ जाती है, जब सूर्य की ऊंचाई का कोण 45º से 30º तक बदल जाता है। टावर की ऊंचाई है –
(A) 5(√3+1) m
(B) 5(√3-1) m
(C) 13 m
(D) 5/(√3+1) m
1
tan30° = a/(a+10)
1/√3 = a/(a+10)
a/√3  = a+10
a/√3 –a = 10
a(√3 -1) = 10
a= 10/(√3 -1) remove root
a= [10/(√3 -1)] × [(√3 +1)/ (√3 +1)]
10×(√3 +10) / (3-1) = 10(√3 +10) / 2
= 5√3 +5
= 5(√3 +1)



112. The value of cos1 º cos2º ……….cos100º  is
Cos1 º cos2º ......... .cos100º का मान है
(A) ∞o
(B) 0
(C) 1
(D) ½
B
cos90° = 0
hence any ratio multiplied by 0 is 0
0-0 = 0

113.  In a town number of maternal death is 50 in a year. If number of infants born in this year was 500, then maternal mortality rate per 1,00,000 is
एक शहर में एक वर्ष में मातृ मृत्यु की संख्या 50 है। यदि इस वर्ष पैदा हुए शिशुओं की संख्या 500 थी, तो 1,00,000 प्रति मातृ मृत्यु दर है
(A)  90,000
(B) 1,000
(C) 10,000
(D) 10,00,000
C
Infants born in one yeat =500 and maternal deaths =50
maternal mortality rate per 1, 00, 000
= 50×1,00,000/500
10,000

114. For the following data, mean deviation from median is
निम्नलिखित डेटा के लिए, औसत से माध्य विचलन है
x = 2, 6, 10, 14, 18
f = 4, 6, 8, 5, 2
(A) 9.2
(B) 0
(C) 3.68
(D) 3.84
x = 2, 6, 10, 14, 18 = Median = n+1/2 th item i.e 3rd item value = 10
f = 4, 6,   8,   5,   2 =  ∑f  = 25
|d| = 8, 4, 0, 4, 8  (diviation of mean from x)
f|d| = 32, 24, 0, 20, 16 = ∑f|d|  = 92
mean deviation from median M.D. = (∑f|d|)/ ∑f 
M.D = 92/25 = 3.68

115. sin21º +sin23º +sin25º +…….sin289º is –
sin21º + sin23º + sin25º + ...... .sin289º  है
(A) 45
(B) 0
(C) 22+1/2
(D) 23
C
In the series 1,3,5……89
Tn =89, a= 1 and d=2
Tn = a+(n-1) d
89 = 1+(n-1)× 2,  89= 1+2n-2, 2n = 89+2-1
2n = 90, n=45 so there will be 45 terms in the series
We know that sin2A+cos2A = 1
taking1st and 45th term  that is sin21 and sin289 (sin289 is equal to cos21)
therefore sin21+cos21 or sin21+sin289 = 1
similarly sin23+cos23 or sin23+sin28 = 1
total terms n=45 and  1 terms of sin21+ sin289 = 1
there will 22 terms upto sin243+ sin247 (sin21 to 43 = 22 and + sin247 to 89 = 22)
the value of 22 terms is 22×1 = 22
23rd term is sin245 , value of sin245 is (1/√2)2 = 1/2
therefore value of sin21 to sin289 = 22+1/2




116. Consider the following pie chart to find the expenditure on Indian Military of India had a total expenditure of 120.
(A) 72 crores 

(B) 66 crores 

(C) 70.8 crores 
(D) 54 crores 
C
Given Military expenses are 59% of total expenditure that is 59×120/100 = 70.8%





  117. What is the angle of elevation of the sun when the length of shadow of a vertical pole is equal to its height?



सूरज की ऊंचाई का कोण क्या होता है जब लंबवत ध्रुव की छाया की लंबाई इसकी ऊंचाई के बराबर होती है?



 


(A) 75º 
(B) 30º


(c) 45º 
(D) 60º

C


length of shadow is Object Height / Sun Tangent = Shadow Length.


tan 45º  = 1/1 = 1





































thus angle of elevation of sum = 45º



 


118. If sin2A = cos(A — 18º), then the value of A is - 

(A) 72 º 

(B) 18 º


(C) 36 º 

(D) 54 º 


C


sin2A = cos(A — 18º   ….(1)


हम जानते हैं की कोस (90-A) = sinA


sin2A = cos(90-2A)


from equation  (1)


cos(90-2A) = cos(A-18)


that means 90-2A = A -18


3A  =  90+18, 3A = 108, A=36 


 



119. If the diameter of a circle is 8 m, which length of arc intercepts a central angle of 45 degree at the centre?



यदि सर्कल का व्यास 8 मीटर है, तो आर्क की लंबाई केंद्र में 45 डिग्री के केंद्रीय कोण को काटती है?(



(A) 4∏ m



(B) ∏ m



(C) 2∏ m



(D) 3∏ m



B



Given : diameter of circle = 8m



radius of circle = 4m



circumfereance of circle 360º= 2∏r



=2∏4



circumfereance of circle 45º



=45×2∏4/360

                          



=∏ m





























































120. In a data, 10 numbers arranged in ascending order. If the 7th observation from the beginning is increased by 4, then the median increase by







डेटा में, 10 संख्या आरोही क्रम में व्यवस्थित की गई। यदि शुरुआत से 7 वां अवलोकन 4 तक बढ़ जाता है, तो औसत बढ़ता है









(A) 5 



(B) 0 



(C) 4



(D) 6



B



Median = (n+1)/2th item that is average of 5th and 6th iten that is 5.5



Since value increase by 4 from 7th observation there will be no change in median







121. The quadratic equation with one root (√5+2) will be 



(A)  x2  - 2√5x + 1 = 0



(B) x2  + 2√5x + 1 = 0



(C) x2  - 4x + 1 = 0



(D) x2  - 4x - 1 = 0



D



x = √5 + 2



x-2 = √5



squaring 



(x -2)2 = 5



x2+4-4x =5



x2-4x -5+4 =  0



x2-4x -1 =  0







122. If the sum of three consecutive multiples of 13 is 390. Then second multiple of 13 is ?







यदि 13 के लगातार तीन गुणांक योग 390 है तो फिर 13 का दूसरा गुणांक है?



(A) 156



(B) 117



(C) 130



(D) 143



C



x+2x+3x = 390



6x = 390, x= 65



x1 =65,x2= 130, x3=195



second multiple is 130







123.  Factorsaition (गुणनखंड) of  [x2 + (a +b+c)x + ab + bc] is 



(A) (x+b) (x+a+b+c)



(B) (x+b) (x+a+c)



(C) (x+a) (x+b+c)



(D) (x+c) (x+a+b)









B



x2 + bx + (a+c)x + ab + bc



x(x+b) + (a+c)x +b(a+c)



x(x+b) + (a+c) (x +b)





(x+b) (x + a + c)
















124. The square of 50005 is - 

(A)2505500025


(B)2500500025 

(C)250050025 

(D)2500050025 


B

(a+b)² = a² + b² + 2ab


a= 50000 b= 5


50000² + 5² + 2(5000×5)


2500000000 + 50000 + 25


= 2500050025


 


125. The value of (857375)1/3 is 

(A) 65 

(B) 95


(C) 85 

(D) 75


B


Split the figure into two parts of three digits


First part is 857, Nearest cube root of 9 is 729


first digit is 9


first digit of second part (375) is 5 five is the only figure whose cube ends with 5 therefore second digit is 5


cube root of 857375 is 95


  


 


126. Raj started a business with 37,500. After some time Umed joined the business with 50,000. 


After one year they shared the profit equally. Find after how many months Umed joined the  business. 


राज ने 37,500 के साथ एक व्यवसाय शुरू किया। कुछ समय बाद उमेद 50,000 के साथ कारोबार में शामिल हो गए। 


एक साल बाद उन्होंने लाभ को समान रूप से साझा किया। उमेद व्यापार में कितने महीने के बाद शामिल होने हुआ।

(A) 8 months 


(B) 3 months 


(C) 9 months 


(D) 4 months


C


 


 माना कि x महीने बाद सामिल हुआ


37500  × 12 = 50000 × x


50000x = 450000

x=9









127 A man bought 3 toffees in a rupee. How many toffees in a rupee must he sell to gain 50%? 

(A) 2.5 


(B) 1


(C) 1.5


(D) 2




D


let us take x toffee to sale in a rupee to gain 50% profit


cost price per toffee = 1/3 rupees


profit on one toffee = 50% of 1/3 i.e 1/6


sale price of 1 toffee to gain 50% profit = (1/3)+ (1/6) = 3/6 or ½ 


½ rupee for 1 toffee


therefore for 1 rupee 1x2 = 2 toffee


 


128. 21% of a fixed number is 28, then 15% of same number is - 

(A) 24 

(B) 18 

(C) 20


(D) 21 


C


x×21/15 = 28


21x/100 = 28


21x=28 so 15x = 28×15/21 = 20






129. The ratio of the length and perimeter of a rectangular plot is 1:3. What is the ratio of the length and breadth of the plot 

(A) 4:3


(B) 1:2 

(C) 2:1


(D) 3:2


C


suppose length is 1 and height is h


then perimeter =  2(1+h) = 3 


2h +2 = 3, 2h=  3-1, 2h = 1


h= ½


Ratio of length and height = 1: 1 /2 = 2:1


 


130. logba x logcb x logac is equal to 

(A) 1 


(B) log abc 

(C) abc 

(D) 1/abc



A



(log a /log b) x (log b /log c) × (log c /log a)



cancelling each other results = 1







131. Rs 4,000 were divided into two parts such a way that first part was invested at 3% and the second at 5% per annum. If the whole simple interest from both the investments for one year is 144. Then the amount of the first part is - 

(A) 2,400 


(B) 2,700 

(c) 2,800 


(D) 2,500 

C


1 st part of amount = x then second part = 4000-x


Total interest on both parts = (3x/100) + [(4000-x)5/100]


= (3x/100) + [(20000-5x)/100] = 144


3x + 20000-5x = 14400


-2x = -5600


x= 2800


 


132 A man spends 75% of his income. His income increases by 20% and his expenditure also increases by 10%, 


then the percentage increase in his savings is - 

(A) 50%


(B) 10%


(C) 20% 

(D) 25% 

A


Income 100 - spends 75 = saving 25


Income + 20% =120,   - spends +10%=82.5, = savings 120-87.5 = 37.5


Savings increase from 25 to 37.5 i.e 50%


 


133 The ratio of an external angle and an internal angle of a regular polygon is 1:5. 


The number Of sides in the polygon is - 
(A) 10 
(B) 12 


(C) 6


(D) 8

B


external angle 1x, internal angle 5x


1x+5x = 6x = 180, x= 30º


so external angle 30 and internal angle = 150º


external angle = 360/ No of sides


30 =  360/ No of sides


Np of sides = 360/30 = 12


 


134. The mean deviation from the mean for the following marks is — 

37, 48, 50, 23, 47, 58, 29, 27, 31, 40 


 (A) 9.8


(B) 9.5


(C) 9.6

(D) 9.7 



C



Convert into cumulative frequency



23, 50, 79, 110,  147, 187, 234,282, 332,390



Mean = ∑f/n, = 360/10 = 39



Deviation from Mean



16, 12, 10, 8, 2, 1, 8, 9, 11, 19 ∑d = 96



Mean Deviation = (∑|d|)/ ∑f  =  96/10 = 9.6






135. The mean Of 13 numbers is 24. If 3 is added to each number, then What will be the new mean ? 

(A) 25 

(B) 24


(C) 21 

(D) 27 

D


Mean of 13 is 24, ∑x = 24×13 = 31


if 3 is added to each 13×3 = 39 new ∑x =312+39 = 351


Mean = 351/13 = 27


 


136. If ∠ABC=∠EFG and AB=EF, then the values of x and y are – 



(A) x=1, y=3



(B) x=1, y=1



(C) x=4, y=1



(D) x=1, y=4



D









































































10x = 2x+8, 10x -2x=8, 8x = 8, 



x=1



10x-y = y+2x,  10-y=y+2, 10-2 = 2y



2y =8,



y =4

                                                                                                                                            










137. The co-ordinate of point p which lie on y-axis and distance of 2 unit from the origin (0, 0) will be - 

(A) (1,1)



(B) (0,2)



(C) (2,0)



(D) (2,2)



B

          



(0,2)



 







 
























138.  If a/3 =b/5 = c/7, then the Value of (a+b+c)b is 


(A) 1/5 

(B) 3


(C) 5


(D) 1/3 

B


a/3 =b/5 = c/7 then 


a=3 , b=5, c=7


(a+b+c)/b  = (3+5+7)/5, 15/5 =3


 


139 Sum of length, breadth and height of a cuboid is 20 cm and the length of its diagonal is 6√6 cm. Its total surface area is - 

(A) 246 cm2 


(B) 216 cm2 
(C) 616 cm2 


(D) 184 cm2 

D


 



Suppose
length, breadth and height as a, b and c respectively therefore



 (a+b+c) =20 



and
diagonal of cuboid √(a²+b²+c²)   = 6√6 or 
√(36×6) = √216, or (6√6)²= 216



So
applying algebraic formula 2(ab+bc+ca) = (a+b+c)²- (a²+b²+c²).



So
2(ab+bc+ca) = 20² - (6√6)² = 400 - 236 = 184 cm²


140. If log10 [log2(log4x)] = 0, then the value of x is – 


(A) 16



(B) 4 



(C) 8



(D) 80



A



Given problem is 



loglO [log2(log4x)
= 0



Now
we have to find the value of x. 



In order to find value of x, we need to isolate x which can be done by converting
logarithmic equation into exponential equation using formula:



logb(a)
= c or bc = a



In this formula, base (b) remains fixed at it's position while other values a and
c switch. So applying this formula on given problem, we get:



l00 = log2(log4(x))




1
= log2(log4(x))




log2(log4(x))  = 1



Apply
above formula again.



21 = (log4(x))



2 = (log4(x))



(log4(x))
= 2



Apply
above formula again.



42
= x



x=
16








141. Mohan bought two cows in 30,000. By selling one at a loss of 15% and other at a gain of 19%, he found that the selling price of both cows is the same. The cost price of both the cows are - 

(A) 20,000, 10,000 

(B) 17,500, 12,500 

(c) 15,000, 15,000 

(D) 18,000, 12,000 

B


Mohan bought first cow in x, then second in 30000-x


1st sold at rate o 85% second @ 119% and selling price of both is same


so 85x = 119(30000-x)


= 85x = 3570000 - 119 x


= 204x = 3570000, or x = 17500


Cost price of !st cow is 17500 and second cowis (30000-17500) = 12500


 


142 The circumference of two circles is in the ratio 2:3, then the ratio of their areas is


(A) 4:9 

(B) 2:3

(c) 3:2

(D) 9:4

A


radius of circle one = r , radius of circle one = R 


ratio of circumference is  2/3 = 2πr/3πR


area of circle= πr2


ratio of area  = πr2 / πR2


= 22 / 32 = 4/9 = 4:9


 


143 A spherical ball of diameter 21 cm is melted and recast into cubes, each of side I cm, then the number of cubes thus formed is - 

(A) 4851 

(B) 2100


(C) 4200 

(D) 4410


A


diameter of ball = 21 cm then radius of ball = 21/2 cm


volume of sphere = 4/3 πr3


= 4/3 × 22/7 × 21/2 × 21/2 × 21/2


= 11×21×21 = 4851 cm3


 


144.  Which of the following pair of equations has no solution ? 

(A) x+2y-5= 0 and 2x+4y-12 = 0

(B) x-2y=0 and 3x+4y-20=O 

(C) x+2Y+1=O and 2x+4Y+2=O 

(D) 2x-y=0 and x-2y=0


A


condition for no solution a1/b2  =  1b/b2  ≠  c1/c2


ax2 + bx + c = 0


let us examine each equation putting values thereof 


(A)  x+2y-5 = 0 and 2x+4y-12 = 0


½ = 2/4 ≠ -5/-12  = No solution


(B) x-2y=0 and 3x+4y-20=O 

1/3 ≠ 3/4 ≠ 0/-20  = There is unique solution


(C) x+2Y+1=O and 2x+4Y+2=O 

½ = 2/4 = 1/2  = There are infinite solutions


(D) 2x-y=0 and x-2y=0


2/1 ≠ -1/*2 ≠ 0/0  = There is unique solution




145. In figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. 


If ∠DBC=55º and ∠BAC45º, then ∠BCD is

(A) 45º 

(B) 100 º 


(C) 80 º 


(D) 35 º 

C






given that ∠CAD = ∠DBC = 55     (Since angles in the same segment)


∠DAB = ∠CAD + ∠BAC


         = 55 + 45         = 100


 


similarly ∠DAB + ∠BCD = 180   (Opposite angle of a cyclic quadrilateral)


 


 = ∠BCD = 180 - ∠DAB


                = 180 - 100  => ∠BCD = 80


 


146. A cylindrical vessel of base radius 20 cm filled up to a height of 32 cm with juice. 


The juice is filled in cylindrical glass of radius 4 cm upto a height of 8 cm. How many such glasses can be filled ? 

(A) 120 


(B) 20

(C) 50 


(D) 100


D 


Volume of cylinder containing juice  = πr² × h


π20² × 31 = ( 22/7) ×20×20×32    …….(1)

Volume of glass containing juice  = πr² × h


π4² × 8 = ( 22/7) ×4×4×8--------(2)


Divide              (1) with (2)


[( 22/7) ×20×20×32] / [( 22/7) ×4×4×8]


after cancelling 5×5×4 = 100





147. If P(9a-2, -b) divides line segment joining the points A(3+1a,-3) and B(8a,5) 


in the ratio 3:1, then the values of a and b are


(A) a=1, b= -3


(B) a=1, b=3


(C) a=-1, b=3


(D) a=-1, b=-3


B


 


9a-2 = [3(8a) +  (3a+1)] / 3+1


4(9a-2) = 24a + 3a + 1


= 36a -27a = 1+8


= 9a=9, a= 1


similarly


b = [3(5) + 1(-3)] / 4


4b = [3(5) + 1(-3)] / 4


4b = 15-3, 4b = 12, b=3


 


148. The two opposite vertices of a square are (-1,2) and (3, 2), then the coordinates of the other two vertices are - 

(A) (-1,0) and (1 -4) 

(B) (1,0) and (1, 4) 

(C) (—1,0) and (-1,4) 

(D) (1,0) and (-1,4) 


B



Let ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinates of vertex B be (x, y).



 AB = BC  (As ABCD is a square)



AB2 = BC2



 [x – (–1)] 2 + (y – 2)2 = (x – 3)2 + (y – 2)2 



(x + 1)2 = (x – 3)2



x 2 + 2x + 1 = x 2 – 6x + 9



 2x + 6x = 9 – 1



8x = 8



x = 1



 In ΔABC, we have:



 AB2 + BC2 = AC2  (Pythagoras theorem)



 2AB2 = AC2  (Since, AB = BC)



 2[(x – (–1))2 + (y – 2)2] = (3 – (–1))2 + (2 – 2)2



 2[(x + 1)2 + (y – 2)2] = (4)2 + (0)2



 2[(1 + 1)2 + (y – 2)2] = 16  ( x = 1)



 2[ 4 + (y – 2)2] = 16



 8 + 2 (y – 2)2 = 16 



2 (y – 2)2 = 16 – 8 = 8



 (y – 2)2 = 4



 y – 2 = ± 2



y – 2 = 2 or y – 2 = –2



 y = 4 or y = 0



 Thus, the other two vertices of the square ABCD are (1, 4) and (1, 0).






 


149 In figure ∠XYZ is produced to point P. If ray YQ bisects ∠ZYP, 


then the value of ∠XYQ is
(A) 302º
(B) 122 º


(C) 105 º 
(D) 116 º 




B






























Value of ∠ XYQ is (180-64)/2  = 116/2  = 58º


∠XYQ = 65 + 58 = 122º


 


150. Successive discounts of p% and q% are equivalent to a discount of 

(A) (p+q /2)%


(B) (p+q)%


(C) [(p+q - (p/100)]%


(D) (p+q-pq/100)%



D







Successive discount = If the first discount is x% and second discount is y$ then



Total discount =[x+y-xy/100]%



In this case [p+q - pq/100%











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