INFOSYS CRYPTARITHMETIC QUESTIONS WITH ANSWERS 2018
Question 1 USA+USSR=PEACE FIND P+E+A+C+E
USA+ USSR= PEACE Here P is carry , P=1 when p=1, E=0 with carry 1 AND U=9 A+R=E=0 with carry 1.so, A=2 and R=8 U+S=A=2 with carry 1, S=3 S+S+1=C, 3+3+1=c=7 932+9338=10270 so,P+E+A+C+E=1+0+2+7+0=10 ANS 10
Question 2 tee+let=All where E=5 find A+L+L
T E E + L E T = A L L SO, suppose E=5 and L =6, then L=1(carry 1), again E +E i.e 5+5+(carry 1)=1, again (carry 1); earlier we have taken T=6 and after summation L= 1 so, T+L=6+1+(carry 1)=8 so, A=8, L=1 A+L+L=8+1+1=10
Question 3 If Ever + Since = Darwin then D + a + r + w + i + n is ?
As it’s a sum of 3 numbers hence the maximum value of D could be 2 or 1, but then we look at S which could have maximum value 9 and if it gets a carry of 1 then the value of A will be 0 and the D=1. 2. Then we looks at E and I which results in R, as 9 is already occupied by S so E could be 8 and I could be 7 and R will be 5 with carry 1. 3. Then we place the values of R=5 and E=8 resulting in N=3 with carry 1. 4. In ten’s digits – E=8 and and value of C could be 3,4,6 (As all others are occupied) and To make I=7 (previously assigned in step 2) E must be added by 9 which is already occupied by S. So this hit ends here. Now start again with reduced value of E or I, repeat above steps until you get a correct answer or a dead end. 5. Repeating above steps one time comes with E=5 and I=7 which results in R=3 carry=1. 6. Placing R=3 and E=5 in unit digits and their sum gives 8. 7. in ten’s digits place E=5 now you can add 2 to make I=7 hence C=2. 8. in 100’s digits you have N=8 (from step 6) and only digits that are left now are 4 and 6 hence. So placing V=4 you will get W=2 which is not true because it is already assigned to C. So we are left with only V=6 as the ultimate choice and this leads the result in to W=4. Hence this way we got the answer ….5 6 5 3 ….E V E R ..9 7 8 2 5 + S I N C E ———————- D A R W I N 1 0 3 4 7 8 Hence D+A+R+W+I+N = 1+0+3+4+7+8 =23
Question 4 If (HE)^H=SHE, where the alphabets take the values from (0-9) & all the alphabets are single digit then find the value of (S+H+E)?
arrange A-J(0-9) likewise K-L(0-9) and U-Z(0-5) the code is S=8,H=7,E=4 so 8+7+4=19 or (HE)^H must be in the range 21 to 29 so that result must be less then 999 (3 digit ,as SHE) .so as per rule sol is (25)^2=625. so S+H+E=6+2+5=13 ANS
Question 5 EAT+EAT+EAT=BEET if t=0 then what will the value of TEE+TEE
EAT+EAT+EAT = BEET. As T=0, no carry for A+A+A(3A). Possible values of A and E can be calculated by 1) 3A= E 2) 3A = 10 +E 3) 3A = 20 + E Here Largest carry generated by addition of three one digit number is 27(9+9+9). Hence value of E is less than 7 for equation 3. For Equation 3) Assume value of E is 7. Therefore value of A=9 now carry + E + E + E = BE. (2) + (7) + (7) + (7) = 23. but 7 is not equal to 3. Contradict to our assumption. Try another value of E as 4 for equation 3 E=4 therefore, A = 8. now carry + E + E + E = BE. (2) + (4) + (4) + (4) = 14. hence value satisfied with our prediction. hence E=4 A=8 and B=1 now TEE + TEE = TAA 044 + 044 = 088 In Above Equations A is integer. So take only those values which are divisible by 3.
Question 6 If CROSS + ROADS = DANGER then D+A+N+G+E+R=?
CROSS= 96233 ROADS=62513 DANGER=158746 Ans 31
Question 7 WORLD+TRADE=CENTER value of C+E+N+T+E+R
WORLD+TRADE=CENTER 53684+ 76042=129726 ANSWER CENTER so,start from last ‘c’ the value is 1 always.Next W+T=E right.By adding numbers we should get 1 as a carry (which is ‘C’ value). so,take W as 5 and T as 7 and add now which is 12. NO TWO ALPHABETS SHOULD HAVE SAME VALUES ALWAYS. Do it accordingly.
Question 8 fine+nine=wives then w+i+v+e+s=?
4 digit + 4digit = 5 digit. fine nine = wives so clearly w=1 as f+n=i+10(it generates a carry over) now,n+n = e,therefore 2n = e. e+e=s, so we can write 2n+2n=s 0r 4n=s. we can see that s is 4 times of n. possible digit between 0-9 that satisfy this condition are 1,4 and 2,8 for n and e respectively. but n cannot be one as the value of w is alreay 1 so the only choice we have is 2,8. therefore n=2 and s=8. from n=2 we can get e=4 now we can put i as 3,5,6,7,9 but if we put i as 5 or above then while finding the value of ” f ” from eqn f+n=i+10 in double digits so i has to be 3. therefore i=3. now i+i=v, so v=6 so finally we get w=1 n=2 e=4 s=8 i=3 v=6 summation of wives =22.
Question 9 USA + USSR = PEACE ; P + E + A + C + E = ?
3 Digit number + 4 digit number = 5 digit number. So P is 1 and U is 9, E is 0. Now S repeated three times, A repeated 2 times. Just give values for S. We can easily get the following table. USA = 932 USSR = 9338 PEACE = 10270 P + E + A + C + E = 1 + 0 + 2 + 7 + 0 = 10
Question 10 If POINT + ZERO = ENERGY, then E + N + E + R + G + Y = ?
4 digit number + 5 digit number = 6 digit number. So E = 1, P = 9, N = 0 Observe R + 0 = G. But R = G not possible. 1 + R = G possible. So R and G are consecutive. G > R. 1 + I = R, So I and R are consecutive. R > I. i.e., G > R > I. and G, R, I are consecutive. Now O + T should give carry over and O + Z also give carry over. So O is bigger number. Now take values for G, R, I as 8, 7, 6 or 7, 6, 5 etc. and do trial and error. POINT = 98504, ZERO = 3168 and ENERGY = 101672. So E + N + E + R + G + Y = 1 + 0 + 1 + 6 + 7 +2 = 17
Question 11 SEND + MORE = MONEY. Then what is the value of M + O + N + E + Y ?
Observe the diagram. M = 1. S + 1 = a two digit number. So S = 1 and O cannot be 1 but 0. Also E and N are consecutive. Do trial and error. SEND = 9567, MORE = 1085, MONEY = 10652 SO M + O + N + E + Y = 1 + 0 + 6 + 5 + 2 = 14