Maths -- Appolo Study Centre
January 14, 2017 | Author: rammi123 | Category: N/A
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Important Formulas: 1. 2. 3. 4. 5. 6. 7. 8.
a 2 − b 2 = (a + b)(a − b) (a + b) 2 = a 2 + b 2 + 2ab (a − b)2 = a 2 + b 2 − 2ab (a + b) 2 = (a − b) 2 + 4ab (a 3 + b3 ) = (a + b)(a 2 − ab + b 2 ) (a 3 − b3 ) = (a − b)(a 2 + ab + b 2 )
(a + b)3 = a 3 + b3 + 3ab(a + b) ⇒ a 3 + b3 = (a + b)3 − 3ab(a + b) (a − b)3 = a 3 − b3 − 3ab(a − b) ⇒ a 3 − b3 = (a − b)3 + 3ab(a − b)
9.
(a + b + c)2 = a 2 + b 2 + c 2 + 2(ab + bc + ca )
1.
Find the value of
343 × 343 × 343 − 113 × 113 ×113 343 × 343 + 343 × 113 + 113 × 113
A. 230
B. 240
C. 235
D. 229
343 × 343 × 343 − 113 × 113 ×113 d; kjpg;G fhz;f 343 × 343 + 343 × 113 + 113 × 113 A. 230
B. 240
C. 235
D. 229
Solution:
a3 − b3 (a − b)(a2 + ab + b2 ) ⇒ ⇒ a − b = 343 −113 = 230 a2 + ab + b2 a2 + ab + b2
PROFIT AND LOSS
Profit = sp – cp
Loss = cp – sp
Profit % =
Loss % =
sp − cp × 100 cp
cp − sp × 100 cp
PERCENTAGE
• Percentage Increase =
Current Value − Pr eviousValue × 100 Pr eviousValue
• Percentage Decrease =
Pr eviosValue − Current Value × 100 Pr eviousValue
1. If the salary of a worker is increased by 20% and then decreased by 10%, what is the percentage effect on his salary? xUtdJ rk;gsk; 20 rjtPjk; mjpfhpf;fpwJ. gpd;G 10 rjtPjk; FiwfpwJ vdpy;> rk;gs rjtPj tpisT vd;d? Solution:
Formula: I – D -
Therefore, 20 – 10 -
ID 100
Note: Increase (I), Decrease (D).
20*10 ⇒ 10 -2 = 8 % increase. 100
2. If the salary of a worker is increased by 20% and then increased by another 10% what is the percentage effect on his salary? xUtdJ rk;gsk; 20 rjtPjk; mjpfhpf;fpwJ. gpd;G NkYk; 10 rjtPjk; mjpfhpf;fpwJ vdpy;> rk;gs rjtPj tpisT vd;d? Formula: I1 + I2 +
Solution:
⇒ 20 + 10 +
I1 I 2 100
20*10 100
= 32% increase.
3. If the price of coffee is increased by 25%, find how much per cent a housewife must reduced her consumption of coffee so as not to increase the expenditure on coffee? fhgpapd; tpiy 25 rjtPjk; mjpfhpf;fpwJ> mth;fsJ FLk;g tuTnrytpy; ve;jtpj khw;wKk; ,y;iy vdpy; fhgp cgNahfpf;Fk; msT vt;tsT rjtPjk; FiwAk;? Solution: Formula:
100r 100 * 25 = ⇒ Ans: 20% 100 + r 100 + 25
4. If the price of wheat falls down by 25%, by how much per cent must a householder increase its consumption, so as not to decrease the expenditure? NfhJikapd; tpiy 25% FiwfpwJ. ,Ug;gpDk;> mth;fsJ FLk;g tuT nrytpy; ve;jtpj khw;wKk; ,y;iy vdpy;> NfhJik cgNahfpf;Fk; msT vt;tsT rjtPjk; mjpfhpf;Fk;? Solution: Formula: Decreased 25%
100r 100* 25 100 1 = = = 33 % 100 − r 75 3 3
SIMPLE INTEREST S .I =
PNR 100
Amount = P + S.I
1. What would be the simple interest obtained on an amount of Rs.8440 at the rate of 15% p.a. after 6 years? &.8440 vd;w njhiff;F Mz;Lf;F 15% tl;b tPjj;jpy; 6 Mz;LfSf;F fpilf;fg;ngWk; jdptl;b njhif vt;tsT? 1. Rs.7596
2. Rs.7648
4. Rs.7816
5. Rs.7888
3. Rs.7720
Explanation SI =
8440 × 6 × 15 = 7596 100
2. At what rate percent per annum simple interest will a sum double itself in 8 years? xU njhifahdJ 8 Mz;Lfspy; ,uz;L klq;fhfpwJ vdpy; jdptl;bapd; tl;b rjtPjk; vt;tsT? Solution: Suppose P = 100, A = 200 S.I.= 200 – 100 = 100; N = 8 years Therefore, R =
S .I × 100 100 × 100 1 = = 12 % P× N 100 × 8 2
COMPOUND INTEREST N
R P 1 + = Amount 100
Where, Amount = P + C.I 1. Calculate the compound interest on a sum of Rs.800 at 5% rate in 2 years. Solution:N
2
R 5 21 21 A = P 1 + = 800 1 + = 800 × × = 882 100 20 20 100 C.I . = 882 − 800 = Rs.82
VISUAL REASONING
Directions: Find the missing character from among the given alternatives. Ex:1
(a)625
(b)25
(c)125
(d)156
Sol: clearly so missing number Hence , the answer is (a).
Ex:2
(a)25
(b)37
(c)41
(d)47
Sol: Clearly in fig In fig In fig (B) , missing number Hence the answer is (c)
.
Ex:3
(a)115
(b)130
(c)135
(d)140
sol: Clearly the number inside the circle is equal to the sum of the product of the upper three numbers and the product of the lower three numbers. Thus, In fig (A) In fig (B) In fig (c), missing answer = Hence the answer is (b). Ex:4
(a)5
(b)4
(c)3
Sol: Clearly in the second column, 22+2-23=1 In the third column, 40+5-43=2 In the first column, missing number = 21+1-20=2 Hence, the answer is (d).
Ex:5
(d)2
(a)11
(b)6
(c)3
(d)2
Sol: Clearly in the first column, In the second column, Let the missing number in the third column be . Then , Hence the answer is (c).
Alpha Numeric Reasoning CODING – DECODING
LETTER CODING: Examples:
1. In a certain code language 'GOAL' is written as 'HPBM'. How will 'POST' be written in that code ? xU re;Njf nkhopapy;, 'GOAL' vDk; thh;ji ; j 'HPBM'vd;W khw;wp vOjg;gLfpwJ. ,ijg;NghyNt 'POST'vDk; thh;ji ; jia vt;thW khw;Wtha;? (A) QPTS
(B) QPST
(C) QPTV
(D) QPTU
Solution:
In the same way
2. In a certain code language 'PINK' is written as 'NGLI’. How will 'IRON' be written in the same code ? xU re;Njf nkhopapy;, 'PINK' vDk; thh;ji ; j 'NGLI’ vd;W khw;wp vOjg;gLfpwJ. ,ijg;NghyNt 'IRON' vDk; thh;ji ; jia vt;thW khw;Wtha;? (A) KPML
(B) GPML
(C) KMPL
(D)GMPE
Solution:
DICE
Example :1.Which symbol will be on the face opposite to the face with symbol * ? * vd;w FwpaPlb ; w;F vjpNu cs;sJ vJ?
a.@
b.$
c. 8
d. +
Answer: C
Explanation: The symbols of the adjacent faces to the face with symbol * are @, -, + and $. Hence the required symbol is 8.
Example :2. Two positions of dice are shown below. How many points will appear on the opposite to the face containing 5 points? ,uz;L ntt;NtW epiyfis nfhz;l xU gfil nfhLf;fg;gl;Ls;sJ. 5 Gs;spfis nfhz;l gf;fj;jpw;F vjpNu vj;jid Gs;spfs; cs;sd?
a.3
b. 1
c. 2
d. 4
Answer: D Explanation: In these two positions one of the common faces having 1 point is in the same position. Therefore according to rule (2). There will be 4 points on the required face.
Example : 3.Which digit will appear on the face opposite to the face with number 4? 4 vd;w gf;fj;jpd; vjpNu cs;s vz; vJ?
a. 3 Answer: A
b. 5
c. 6
d. 2/3
Explanation: Here the common faces with number 3, are in same positions. Hence 6 is opposite to 2 and 5 is opposite to 1. Therefore 4 is opposite to 3.
Example : 4. Two positions of a dice are shown below. Which number will appear on the face opposite to the face with the number 5? 5 vd;w gf;fj;jpw;F vjpNu cs;s vz; vJ?
a. 2/6 Answer: C
b. 2
c. 6
d. 4
Explanation: According to the rule no. (3), common faces with number 3, are in same positions. Hence the number of the opposite face to face with number 5 will be 6.
Example :5. How many points will be on the face opposite to in face which contains 2 points? ,uz;L Gs;spfSf;F vjpNu vj;jid Gs;spfs; cs;sd?
a. 1
b. 5
c. 4
d. 6
Answer: D Explanation: In first two positions of dice one common face containing 5 is same. Therefore according to rule no. (3) the face opposite to the face which contains 2 point, will contains 6 points.
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